Trigonometry

1. Sine rule


ID is: 3551 Seed is: 8804

The sine rule: finding mistakes

In maths class you are learning about the sine rule. You are working on a problem in which the triangle has sides 4, 6, and 7, and two angles labelled A and B. You write down the sine rule equation for this triangle, as shown below. But your teacher walks by and tells you that there is a mistake in the equation.

Incorrect equation:

sinB7=sinA4

What should you do to correct the equation?

Answer:

To correct the equation you should .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule includes one side of the triangle and one angle. That side and angle must be opposite to each other in the triangle. The equation in this question includes the angles A and B. Are the sides in the equation both opposites of A and B?


STEP: Find the sides opposite to angles A and B
[−1 point ⇒ 0 / 1 points left]

The sine rule is an equation which connects two sides and two angles in any triangle:

asinA=bsinB or sinAa=sinBb

where:

  • A and B are two angles in the triangle
  • a is the side of the triangle opposite to angle A
  • b is the side of the triangle opposite to angle B

For this question, that means these two sides are the important ones:

Note that 4 is not relevant because it is not opposite to either of the angles we know. And that is why the equation in this question is wrong:

sinB7=sinA4_

All four of these equations are correct for this triangle:

6sinA=7sinB or sinA6=sinB7 or 7sinB=6sinA or sinB7=sinA6
TIP: The equations above show that the sine rule is quite flexible. Keep this in mind when you are solving problems with the sine rule. You can simplify your work by choosing a more convenient arrangement.

To correct the equation, change the 4 to 6.


Submit your answer as:

ID is: 3551 Seed is: 9182

The sine rule: finding mistakes

In maths class you are learning about the sine rule. You are working on a problem in which the triangle has sides 4, 5, and 6, and two angles labelled A and B. You write down the sine rule equation for this triangle, as shown below. But your teacher walks by and tells you that there is a mistake in the equation.

Incorrect equation:

sinB6=sinA5

What should you do to correct the equation?

Answer:

To correct the equation you should .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule includes one side of the triangle and one angle. That side and angle must be opposite to each other in the triangle. The equation in this question includes the angles A and B. Are the sides in the equation both opposites of A and B?


STEP: Find the sides opposite to angles A and B
[−1 point ⇒ 0 / 1 points left]

The sine rule is an equation which connects two sides and two angles in any triangle:

asinA=bsinB or sinAa=sinBb

where:

  • A and B are two angles in the triangle
  • a is the side of the triangle opposite to angle A
  • b is the side of the triangle opposite to angle B

For this question, that means these two sides are the important ones:

Note that 6 is not relevant because it is not opposite to either of the angles we know. And that is why the equation in this question is wrong:

sinB6_=sinA5

All four of these equations are correct for this triangle:

5sinA=4sinB or sinA5=sinB4 or 4sinB=5sinA or sinB4=sinA5
TIP: The equations above show that the sine rule is quite flexible. Keep this in mind when you are solving problems with the sine rule. You can simplify your work by choosing a more convenient arrangement.

To correct the equation, change the 6 to 4.


Submit your answer as:

ID is: 3551 Seed is: 1033

The sine rule: finding mistakes

In maths class you are learning about the sine rule. You are working on a problem in which the triangle has sides 6, 7, and 9, and two angles labelled A and B. You write down the sine rule equation for this triangle, as shown below. But your teacher walks by and tells you that there is a mistake in the equation.

Incorrect equation:

sinB9=sinA7

What should you do to correct the equation?

Answer:

To correct the equation you should .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule includes one side of the triangle and one angle. That side and angle must be opposite to each other in the triangle. The equation in this question includes the angles A and B. Are the sides in the equation both opposites of A and B?


STEP: Find the sides opposite to angles A and B
[−1 point ⇒ 0 / 1 points left]

The sine rule is an equation which connects two sides and two angles in any triangle:

asinA=bsinB or sinAa=sinBb

where:

  • A and B are two angles in the triangle
  • a is the side of the triangle opposite to angle A
  • b is the side of the triangle opposite to angle B

For this question, that means these two sides are the important ones:

Note that 7 is not relevant because it is not opposite to either of the angles we know. And that is why the equation in this question is wrong:

sinB9=sinA7_

All four of these equations are correct for this triangle:

6sinA=9sinB or sinA6=sinB9 or 9sinB=6sinA or sinB9=sinA6
TIP: The equations above show that the sine rule is quite flexible. Keep this in mind when you are solving problems with the sine rule. You can simplify your work by choosing a more convenient arrangement.

To correct the equation, change the 7 to 6.


Submit your answer as:

ID is: 1556 Seed is: 2171

Using the sine rule

Given a triangle with an angle of 79° and two sides with lengths 8.1 and 12.8, find the missing angle, indicated by the question mark in the figure. The figure below is drawn to scale.

TIP: You can compare the answer you get to the diagram because it is drawn to scale. The angle that you want is acute, so check your answer before you submit it to make sure it agrees with the figure.
INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.
Answer: The missing angle is °
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula and go for it!


STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 4 / 5 points left]

The sine rule works for any triangle, and it relates two angles to the two sides opposite to those angles. We can use the sine rule in either of two ways:

Angles in numerators:sinA^a=sinB^bSides in numerators:asinA^=bsinB^

Both versions will work, but we should pick the version that makes it easier to solve the question. And in this case the first version above is the better choice because we want to find an angle and it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So it will be easier to use this arrangement of the formula:

sinA^a=sinB^b

STEP: Substitute in the known values
[−1 point ⇒ 3 / 5 points left]

When using the sine rule, we must substitute in the values carefully to make sure that they are in the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Notice that we also labelled the vertices of the triangle, in case that will be helpful. In fact, the angle we want is C^. We can use that in our equation:

sinC^8.1=sin(79°)12.8

STEP: Solve for the missing angle
[−3 points ⇒ 0 / 5 points left]

Now it is a matter of solving for C^.

sinC^8.1=sin(79°)12.8sinC^8.1=(0.98162...)12.8sinC^=(0.07668...)(8.1)sinC^=0.62118...
At this point we must use the inverse sine operation. Whenever we do that, we must remember the second group of solutions and the +360°k business. The good news is that we can ignore the +360°k term now: we know the angle must be between zero and 180° because it is in a triangle. However, we cannot ignore the second solution: we must figure out if it gives us a real answer or not. (Remember that the second solution for the sine function comes from 180θ, where θ is whatever angle you get from your calculator.)
C^=sin1(0.62118...)C^=38.40279...°or141.59720...°

Now we must figure out which of these answers is correct. Or are they both possible? It turns out that the first answer, C^38.4°, is the only possible answer. We know this by comparing to the diagram: the angle we want is an acute angle (less than 90°). So while 141.6° is a valid solution to the equation we solved, it is not a valid angle for the triangle in this question. (In this case, there is another problem with 141.6°: if we add it to the angle given in the question, we get 141.6°+79°=220.6°. And the angles in a triangle cannot be larger than 180°.)

NOTE: It is possible that a question like this will have two solutions, or that the obtuse angle is the correct answer. For example, if the diagram was not drawn to scale, we could not assume that the angle is acute. You should not assume that the answer will always be an acute angle!

The instructions said to round to one decimal place, so the final answer is C^=38.4°.


Submit your answer as:

ID is: 1556 Seed is: 9863

Using the sine rule

Given a triangle with a known side of 6.9 and two known angles of 54° and 60°, find the length of the side labelled with the question mark. The figure below is drawn to scale.

TIP: You can compare the answer you get to the diagram because it is drawn to scale. The side you need is a similar length to the given side 6.9 so check your answer before you submit it to make sure it agrees with the figure.
INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.
Answer: The length of the missing side is units.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula and go for it!


STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 3 / 4 points left]

The sine rule works for any triangle, and it relates two angles to the two sides opposite to those angles. We can use the sine rule in either of two ways:

Angles in numerators:sinA^a=sinB^bSides in numerators:asinA^=bsinB^

Both versions will work, but we should pick the version that makes it easier to solve the question. And in this case the second version above is the better choice because we want to find a side and it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So it will be easier to use this arrangement of the formula:

asinA^=bsinB^

STEP: Substitute in the known values
[−1 point ⇒ 2 / 4 points left]

When using the sine rule, we must substitute in the values carefully to make sure that they are in the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Notice that we also labelled the vertices of the triangle, in case that will be helpful. Each pair belongs together in one fraction. Let's use x to represent the side we are looking for (the one with the '?').

xsin(54°)=6.9sin(60°)

STEP: Solve for the missing side
[−2 points ⇒ 0 / 4 points left]

Now it is a matter of solving for x.

xsin(54°)=6.9sin(60°)xsin(54°)=6.9(0.86602...)
Now work on the value of the fraction on the right side and multiply both sides of the equation by the sin(54°) to cancel the denominator on the left.
x=(7.96743...)sin(54°)x=6.44578...x6.4

Presto! We have the answer. The final step rounds the answer to one decimal place, as required in the instructions.

The unknown side of the triangle has a length of 6.4 units.


Submit your answer as:

ID is: 1556 Seed is: 4076

Using the sine rule

Given a triangle with an angle of 85° and two sides with lengths 7.5 and 13.9, find the missing angle, indicated by the question mark in the figure. The figure below is drawn to scale.

TIP: You can compare the answer you get to the diagram because it is drawn to scale. The angle that you want is acute, so check your answer before you submit it to make sure it agrees with the figure.
INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.
Answer: The missing angle is °
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula and go for it!


STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 4 / 5 points left]

The sine rule works for any triangle, and it relates two angles to the two sides opposite to those angles. We can use the sine rule in either of two ways:

Angles in numerators:sinA^a=sinB^bSides in numerators:asinA^=bsinB^

Both versions will work, but we should pick the version that makes it easier to solve the question. And in this case the first version above is the better choice because we want to find an angle and it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So it will be easier to use this arrangement of the formula:

sinA^a=sinB^b

STEP: Substitute in the known values
[−1 point ⇒ 3 / 5 points left]

When using the sine rule, we must substitute in the values carefully to make sure that they are in the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Notice that we also labelled the vertices of the triangle, in case that will be helpful. In fact, the angle we want is C^. We can use that in our equation:

sinC^7.5=sin(85°)13.9

STEP: Solve for the missing angle
[−3 points ⇒ 0 / 5 points left]

Now it is a matter of solving for C^.

sinC^7.5=sin(85°)13.9sinC^7.5=(0.99619...)13.9sinC^=(0.07166...)(7.5)sinC^=0.53751...
At this point we must use the inverse sine operation. Whenever we do that, we must remember the second group of solutions and the +360°k business. The good news is that we can ignore the +360°k term now: we know the angle must be between zero and 180° because it is in a triangle. However, we cannot ignore the second solution: we must figure out if it gives us a real answer or not. (Remember that the second solution for the sine function comes from 180θ, where θ is whatever angle you get from your calculator.)
C^=sin1(0.53751...)C^=32.51464...°or147.48535...°

Now we must figure out which of these answers is correct. Or are they both possible? It turns out that the first answer, C^32.5°, is the only possible answer. We know this by comparing to the diagram: the angle we want is an acute angle (less than 90°). So while 147.5° is a valid solution to the equation we solved, it is not a valid angle for the triangle in this question. (In this case, there is another problem with 147.5°: if we add it to the angle given in the question, we get 147.5°+85°=232.5°. And the angles in a triangle cannot be larger than 180°.)

NOTE: It is possible that a question like this will have two solutions, or that the obtuse angle is the correct answer. For example, if the diagram was not drawn to scale, we could not assume that the angle is acute. You should not assume that the answer will always be an acute angle!

The instructions said to round to one decimal place, so the final answer is C^=32.5°.


Submit your answer as:

ID is: 3556 Seed is: 5160

Working algebraically: finding a sine ratio

Triangle BAC is shown below with sides labelled 3b, 20b48b3, and 5b2. All three of these expression are non-zero. A is labelled θ. The diagram is not drawn to scale.

  1. Determine an expression for sinθ in terms of b and sinC.

    Answer: sinθ=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The answer should be "in terms of b and sinC". That means the answer will include both the variable b and the expression sinC. You can get both of these into an equation using the sine rule.


    STEP: Set up an equation using sinC and b
    [−1 point ⇒ 2 / 3 points left]

    The question says that we need an answer "in terms of b and sinC". That means we want an expression which includes both b and the expression sinC. How can we do that? We can use the sine rule!

    sinθ20b48b3=sinC5b2

    STEP: Solve for sinθ
    [−1 point ⇒ 1 / 3 points left]

    We want an expression for sinθ. So we need to rearrange the equation above to make sinθ the subject. To do this, multiply both sides of the equation by 20b48b3.

    sinθ20b48b3=sinC5b2sinθ=sinC5b2(20b48b3)

    STEP: Simplify the expression as much as possible
    [−1 point ⇒ 0 / 3 points left]

    The result above agrees with the question: it is an expression for sinθ in terms of both b and sinC. But we can simplify the expression on the right side: we can factorise the numerator. We can take out the common factor 4b3.

    sinθ=sinC5b2(20b48b3)=(20b48b3)sinC5b2=(4b3)(5b2)sinC5b2=(4b3)sinC

    And there we have it: the expression for sinθ in terms of b and sinC is (4b3)sinC.


    Submit your answer as:
  2. Question 1 states that the three side expressions are non-zero. Why must those expressions be non-zero?

    Answer:

    The expressions must be non-zero .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Any time there are fractions involved in a calculation, you must worry about division by zero. Why is that?


    STEP: Select the correct answer
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why the side expressions for the triangle must be non-zero. The reason is a common one: whenever there are fractions, we must be sure not to divide by zero. The solution to Question 1 uses the sine rule, and that includes fractions. So the statement "All three of these expression are non-zero" prevents division by zero. And that prevents undefined fractions.

    The correct answer is that the expression must be non-zero to avoid undefined fractions.


    Submit your answer as:

ID is: 3556 Seed is: 3754

Working algebraically: finding a sine ratio

Triangle ABC is shown below with sides labelled 3y2, 15y5, and 5y. All three of these expression are non-zero. B is labelled θ. The diagram is not drawn to scale.

  1. Determine an expression for sinθ in terms of y and sinA.

    Answer: sinθ=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The answer should be "in terms of y and sinA". That means the answer will include both the variable y and the expression sinA. You can get both of these into an equation using the sine rule.


    STEP: Set up an equation using sinA and y
    [−1 point ⇒ 2 / 3 points left]

    The question says that we need an answer "in terms of y and sinA". That means we want an expression which includes both y and the expression sinA. How can we do that? We can use the sine rule!

    sinθ15y5=sinA5y

    STEP: Solve for sinθ
    [−1 point ⇒ 1 / 3 points left]

    We want an expression for sinθ. So we need to rearrange the equation above to make sinθ the subject. To do this, multiply both sides of the equation by 15y5.

    sinθ15y5=sinA5ysinθ=sinA5y(15y5)

    STEP: Simplify the expression as much as possible
    [−1 point ⇒ 0 / 3 points left]

    The result above agrees with the question: it is an expression for sinθ in terms of both y and sinA. But we can simplify the expression on the right side: we can factorise the numerator. In fact, we can split the numerator such that we get a factor which matches the denominator.

    sinθ=sinA5y(15y5)=(15y5)sinA5y=(3y4)(5y)sinA5y=(3y4)sinA

    And there we have it: the expression for sinθ in terms of y and sinA is (3y4)sinA.


    Submit your answer as:
  2. Question 1 states that the three side expressions are non-zero. Why must those expressions be non-zero?

    Answer:

    The expressions must be non-zero .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Any time there are fractions involved in a calculation, you must worry about division by zero. Why is that?


    STEP: Select the correct answer
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why the side expressions for the triangle must be non-zero. The reason is a common one: whenever there are fractions, we must be sure not to divide by zero. The solution to Question 1 uses the sine rule, and that includes fractions. So the statement "All three of these expression are non-zero" prevents division by zero. And that prevents undefined fractions.

    The correct answer is that the expression must be non-zero to avoid undefined fractions.


    Submit your answer as:

ID is: 3556 Seed is: 7904

Working algebraically: finding a sine ratio

Triangle CAB is shown below with sides labelled 3a2, 9a4, and 3a4. All three of these expression are non-zero. A is labelled α. The diagram is not drawn to scale.

  1. Determine an expression for sinα in terms of a and sinC.

    Answer: sinα=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The answer should be "in terms of a and sinC". That means the answer will include both the variable a and the expression sinC. You can get both of these into an equation using the sine rule.


    STEP: Set up an equation using sinC and a
    [−1 point ⇒ 2 / 3 points left]

    The question says that we need an answer "in terms of a and sinC". That means we want an expression which includes both a and the expression sinC. How can we do that? We can use the sine rule!

    sinα9a4=sinC3a2

    STEP: Solve for sinα
    [−1 point ⇒ 1 / 3 points left]

    We want an expression for sinα. So we need to rearrange the equation above to make sinα the subject. To do this, multiply both sides of the equation by 9a4.

    sinα9a4=sinC3a2sinα=sinC3a2(9a4)

    STEP: Simplify the expression as much as possible
    [−1 point ⇒ 0 / 3 points left]

    The result above agrees with the question: it is an expression for sinα in terms of both a and sinC. But we can simplify the expression on the right side: we can factorise the numerator. In fact, we can split the numerator such that we get a factor which matches the denominator.

    sinα=sinC3a2(9a4)=(9a4)sinC3a2=(3a2)(3a2)sinC3a2=(3a2)sinC

    And there we have it: the expression for sinα in terms of a and sinC is (3a2)sinC.


    Submit your answer as:
  2. Question 1 states that the three side expressions are non-zero. Why must those expressions be non-zero?

    Answer:

    The expressions must be non-zero .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Any time there are fractions involved in a calculation, you must worry about division by zero. Why is that?


    STEP: Select the correct answer
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the reason why the side expressions for the triangle must be non-zero. The reason is a common one: whenever there are fractions, we must be sure not to divide by zero. The solution to Question 1 uses the sine rule, and that includes fractions. So the statement "All three of these expression are non-zero" prevents division by zero. And that prevents undefined fractions.

    The correct answer is that the expression must be non-zero to avoid undefined fractions.


    Submit your answer as:

ID is: 3569 Seed is: 2450

Finding an angle with the sine rule

In ΔPQR, R^=99°, RP=6.9 cm, and PQ=9 cm.

Calculate the size of Q^.

INSTRUCTION: Round the answer to two decimal places.
Answer: Q^= °
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation using the sine rule. The equation should look like this:

sinQ^a side=sin(an angle)another side

STEP: Write an equation using the sine rule
[−1 point ⇒ 2 / 3 points left]

We have a triangle with two sides and one angle labelled. And we are looking for an angle. This is a sine rule situation, because the sine rule always includes two sides and two angles in a triangle.

TIP:

There are two ways we can write the sine rule. Choose the way that makes it the easiest to solve the question:

  • Angles in numerators and sides in denominators: sinA^a=sinB^b
  • Sides in numerators and angles in denominators: asinA^=bsinB^

Write the equation so that whatever you need to solve for is in the numerator.

Using the given information, we can write:

sinA^a=sinB^bsinQ^(6.9)=sin(99°)(9)

STEP: Solve for sinQ^
[−1 point ⇒ 1 / 3 points left]

The first step in solving for the angle is to multiply both sides of the equation by 6.9. This will isolate sinQ^. Then evaluate the expression on the right side.

sinQ^(6.9)=sin(99°)(9)sinQ^=sin(99°)(9)6.9sinQ^=0.75722...

STEP: Use the inverse sine function to find the angle
[−1 point ⇒ 0 / 3 points left]

Now we need to use the inverse sine function to isolate the angle. (Inverses undo each other, so the inverse sine function will take the sine function away from angle Q^.) As always, we do this on both sides of the equation. Then use a calculator to evaluate the result.

sinQ^=0.75722...sin1(sinQ^)=sin1(0.75722...)Q^=49.22040...Q^49.22°

From our calculators we get one answer. However, the inverse sine function provides two more things that the calculator does not show us! Firstly, we get a second answer, which is 180°49.22°. Secondly, the sine curve repeats every 360°. That means we actually get these solutions to the equation:

Q^=49.22°+360°k,kZQ^=130.78°+360°k,kZ

These are all the answers to the equation (there are an infinite number of them). But this question is about an angle in a triangle. This answer represents something real so the answer needs to be realistic. That means that we can ignore the +360°k terms because the answer must be between 0° and 180° (angles cannot be negative and angles in a triangle have a sum of 180°). Additionally, there is already an angle of 99° in the triangle, so angle Q^ must be less than 90° (because the sum of all three angles is 180°). Therefore, the only answer which is acceptable is Q^=49.22°.

The size of Q^ is 49.22°.


Submit your answer as:

ID is: 3569 Seed is: 6626

Finding an angle with the sine rule

In ΔPRQ, Q^=96°, QP=4.9 cm, and PR=6.6 cm.

Calculate the size of R^.

INSTRUCTION: Round the answer to two decimal places.
Answer: R^= °
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation using the sine rule. The equation should look like this:

sinR^a side=sin(an angle)another side

STEP: Write an equation using the sine rule
[−1 point ⇒ 2 / 3 points left]

We have a triangle with two sides and one angle labelled. And we are looking for an angle. This is a sine rule situation, because the sine rule always includes two sides and two angles in a triangle.

TIP:

There are two ways we can write the sine rule. Choose the way that makes it the easiest to solve the question:

  • Angles in numerators and sides in denominators: sinA^a=sinB^b
  • Sides in numerators and angles in denominators: asinA^=bsinB^

Write the equation so that whatever you need to solve for is in the numerator.

Using the given information, we can write:

sinA^a=sinB^bsinR^(4.9)=sin(96°)(6.6)

STEP: Solve for sinR^
[−1 point ⇒ 1 / 3 points left]

The first step in solving for the angle is to multiply both sides of the equation by 4.9. This will isolate sinR^. Then evaluate the expression on the right side.

sinR^(4.9)=sin(96°)(6.6)sinR^=sin(96°)(6.6)4.9sinR^=0.73835...

STEP: Use the inverse sine function to find the angle
[−1 point ⇒ 0 / 3 points left]

Now we need to use the inverse sine function to isolate the angle. (Inverses undo each other, so the inverse sine function will take the sine function away from angle R^.) As always, we do this on both sides of the equation. Then use a calculator to evaluate the result.

sinR^=0.73835...sin1(sinR^)=sin1(0.73835...)R^=47.59165...R^47.59°

From our calculators we get one answer. However, the inverse sine function provides two more things that the calculator does not show us! Firstly, we get a second answer, which is 180°47.59°. Secondly, the sine curve repeats every 360°. That means we actually get these solutions to the equation:

R^=47.59°+360°k,kZR^=132.41°+360°k,kZ

These are all the answers to the equation (there are an infinite number of them). But this question is about an angle in a triangle. This answer represents something real so the answer needs to be realistic. That means that we can ignore the +360°k terms because the answer must be between 0° and 180° (angles cannot be negative and angles in a triangle have a sum of 180°). Additionally, there is already an angle of 96° in the triangle, so angle R^ must be less than 90° (because the sum of all three angles is 180°). Therefore, the only answer which is acceptable is R^=47.59°.

The size of R^ is 47.59°.


Submit your answer as:

ID is: 3569 Seed is: 2714

Finding an angle with the sine rule

In ΔQPR, R^=111°, RQ=3.5 cm, and QP=7 cm.

Calculate the size of P^.

INSTRUCTION: Round the answer to two decimal places.
Answer: P^= °
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation using the sine rule. The equation should look like this:

sinP^a side=sin(an angle)another side

STEP: Write an equation using the sine rule
[−1 point ⇒ 2 / 3 points left]

We have a triangle with two sides and one angle labelled. And we are looking for an angle. This is a sine rule situation, because the sine rule always includes two sides and two angles in a triangle.

TIP:

There are two ways we can write the sine rule. Choose the way that makes it the easiest to solve the question:

  • Angles in numerators and sides in denominators: sinA^a=sinB^b
  • Sides in numerators and angles in denominators: asinA^=bsinB^

Write the equation so that whatever you need to solve for is in the numerator.

Using the given information, we can write:

sinA^a=sinB^bsinP^(3.5)=sin(111°)(7)

STEP: Solve for sinP^
[−1 point ⇒ 1 / 3 points left]

The first step in solving for the angle is to multiply both sides of the equation by 3.5. This will isolate sinP^. Then evaluate the expression on the right side.

sinP^(3.5)=sin(111°)(7)sinP^=sin(111°)(7)3.5sinP^=0.46679...

STEP: Use the inverse sine function to find the angle
[−1 point ⇒ 0 / 3 points left]

Now we need to use the inverse sine function to isolate the angle. (Inverses undo each other, so the inverse sine function will take the sine function away from angle P^.) As always, we do this on both sides of the equation. Then use a calculator to evaluate the result.

sinP^=0.46679...sin1(sinP^)=sin1(0.46679...)P^=27.82614...P^27.83°

From our calculators we get one answer. However, the inverse sine function provides two more things that the calculator does not show us! Firstly, we get a second answer, which is 180°27.83°. Secondly, the sine curve repeats every 360°. That means we actually get these solutions to the equation:

P^=27.83°+360°k,kZP^=152.17°+360°k,kZ

These are all the answers to the equation (there are an infinite number of them). But this question is about an angle in a triangle. This answer represents something real so the answer needs to be realistic. That means that we can ignore the +360°k terms because the answer must be between 0° and 180° (angles cannot be negative and angles in a triangle have a sum of 180°). Additionally, there is already an angle of 111° in the triangle, so angle P^ must be less than 90° (because the sum of all three angles is 180°). Therefore, the only answer which is acceptable is P^=27.83°.

The size of P^ is 27.83°.


Submit your answer as:

ID is: 3500 Seed is: 1083

The flexibility of the sine rule

The diagram below shows a triangle, drawn to scale. One of the sides has a length of 8.6 and another side has the length 9.9. The angle across from the side labelled 8.6 is 45°.

  1. Suppose you need to calculate x, the angle across from the side with length 9.9. The two equations below are both accurate and you can use either of these versions to find x. But one choice makes the solution a bit easier. Which version is the better choice?

    Version 1 9.9sinx=8.6sin(45°)
    Version 2 sinx9.9=sin(45°)8.6
    Answer: The better choice is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think about solving equations these two equations for y:

    y5=710or5y=107

    The equation on the left is more convenient to solve. Why?


    STEP: Choose the version with the needed value in the numerator
    [−1 point ⇒ 0 / 1 points left]

    We can arrange the sine rule in two different ways. We can either put the angles in the numerator with the sides in the denominator, or the other way around:

    sides in denominators:Angles in numerators andsinA^a=sinB^bangles in denominators:Sides in numerators andasinA^=bsinB^

    In both equations, a and b represent sides of the triangle and A^ and B^ represent angles. Solving equations with the unknown value in the numerator is usually easier than solving equations with the unknown in the denominator. So we should use the arrangement which will put the value we want to find in the numerator.

    In this question, we have two versions of the sine rule for the triangle given. Both of these versions are correct. But since we want to find an angle the choice with the angles in the numerators is the better choice.

    The correct choice is Version 2.


    Submit your answer as:
  2. Now solve for the missing angle, x. Use the equation identified in Question 1.

    INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.
    Answer: The missing angle is °
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start with the equation we identified in Question 1:

    sinx9.9=sin(45°)8.6

    You need to solve for x.


    STEP: Solve for the missing angle
    [−3 points ⇒ 0 / 3 points left]

    We need to find the value of x, which is the missing angle. We already have the equation from Question 1. So let's solve it! Start by evaluating the right side of the equation.

    sinx9.9=sin(45°)8.6sinx9.9=(0.70710...)8.6sinx9.9=0.08222...sinx=(0.08222...)(9.9)sinx=0.81399...
    Now we must use the inverse sine operation. This comes with a second group of solutions and a +360°k term. But there is good news: we can ignore the +360°k term. This is because we know the angle must be between zero and 180° because it is in a triangle. However, we cannot ignore the second solution: we must figure out if it gives us a real answer or not. (Remember that the second solution for the sine function comes from 180θ, where θ is whatever angle you get from your calculator.)
    x=sin1(0.81399...)x=54.48811...°or125.51188...°

    We have two answers and we need to figure out which of these answers is correct. Or are they both correct? It turns out that the second answer, x125.5°, is impossible. We know this by comparing to the diagram: the angle we want is an acute angle (less than 90°). So while 125.5° solves the equation, it is not a valid angle for the triangle in this question.

    NOTE: It is possible that a question like this will have two solutions, or that the obtuse angle is the only correct answer. For example, if the diagram was not drawn to scale, we could not assume that the angle is acute. You should not assume that the answer will always be an acute angle!

    Rounded to the tenths place, the angle has a value of x=54.5°.


    Submit your answer as:

ID is: 3500 Seed is: 3449

The flexibility of the sine rule

The diagram below shows a triangle, drawn to scale. Two angles are shown, with values of 37° and 103°. The side of the triangle across from the 103 angle has a length of 9.8 units.

  1. Suppose you need to calculate x, the side across from the 37 angle. The two equations below are both accurate and you can use either of these versions to find x. But one choice makes the solution a bit easier. Which version is the better choice?

    Version 1 sin(37°)x=sin(103°)9.8
    Version 2 xsin(37°)=9.8sin(103°)
    Answer: The better choice is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think about solving equations these two equations for y:

    y5=710or5y=107

    The equation on the left is more convenient to solve. Why?


    STEP: Choose the version with the needed value in the numerator
    [−1 point ⇒ 0 / 1 points left]

    We can arrange the sine rule in two different ways. We can either put the angles in the numerator with the sides in the denominator, or the other way around:

    sides in denominators:Angles in numerators andsinA^a=sinB^bangles in denominators:Sides in numerators andasinA^=bsinB^

    In both equations, a and b represent sides of the triangle and A^ and B^ represent angles. Solving equations with the unknown value in the numerator is usually easier than solving equations with the unknown in the denominator. So we should use the arrangement which will put the value we want to find in the numerator.

    In this question, we have two versions of the sine rule for the triangle given. Both of these versions are correct. But since we want to find a side the choice with the sides in the numerators is the better choice.

    The correct choice is Version 2.


    Submit your answer as:
  2. Now solve for the missing side, x. Use the equation identified in Question 1.

    INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.
    Answer: The length of the missing side is units.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start with the equation we identified in Question 1:

    xsin(37°)=9.8sin(103°)

    You need to solve for x.


    STEP: Solve for the missing side
    [−2 points ⇒ 0 / 2 points left]

    We need to find the value of x, which is the missing side. We already have the equation from Question 1. So let's solve it! Start by evaluating the right side of the equation.

    xsin(37°)=9.8sin(103°)xsin(37°)=9.8(0.97437...)xsin(37°)=10.05778...
    Now work on the value of the fraction on the right side and multiply both sides of the equation by the sin(37°) to cancel the denominator on the left.
    x=(10.05778...)sin(37°)x=6.05292...x6.1

    And that's the answer. The final step rounds the answer to one decimal place, as required in the instructions.

    The unknown side of the triangle has a length of 6.1.


    Submit your answer as:

ID is: 3500 Seed is: 2126

The flexibility of the sine rule

The diagram below shows a triangle, drawn to scale. One of the sides has a length of 8.2 and another side has the length 14.7. The angle across from the side labelled 14.7 is 91°.

  1. Suppose you need to calculate x, the angle across from the side with length 8.2. The two equations below are both accurate and you can use either of these versions to find x. But one choice makes the solution a bit easier. Which version is the better choice?

    Version 1 sinx8.2=sin(91°)14.7
    Version 2 8.2sinx=14.7sin(91°)
    Answer: The better choice is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think about solving equations these two equations for y:

    y5=710or5y=107

    The equation on the left is more convenient to solve. Why?


    STEP: Choose the version with the needed value in the numerator
    [−1 point ⇒ 0 / 1 points left]

    We can arrange the sine rule in two different ways. We can either put the angles in the numerator with the sides in the denominator, or the other way around:

    sides in denominators:Angles in numerators andsinA^a=sinB^bangles in denominators:Sides in numerators andasinA^=bsinB^

    In both equations, a and b represent sides of the triangle and A^ and B^ represent angles. Solving equations with the unknown value in the numerator is usually easier than solving equations with the unknown in the denominator. So we should use the arrangement which will put the value we want to find in the numerator.

    In this question, we have two versions of the sine rule for the triangle given. Both of these versions are correct. But since we want to find an angle the choice with the angles in the numerators is the better choice.

    The correct choice is Version 1.


    Submit your answer as:
  2. Now solve for the missing angle, x. Use the equation identified in Question 1.

    INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.
    Answer: The missing angle is °
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start with the equation we identified in Question 1:

    sinx8.2=sin(91°)14.7

    You need to solve for x.


    STEP: Solve for the missing angle
    [−3 points ⇒ 0 / 3 points left]

    We need to find the value of x, which is the missing angle. We already have the equation from Question 1. So let's solve it! Start by evaluating the right side of the equation.

    sinx8.2=sin(91°)14.7sinx8.2=(0.99984...)14.7sinx8.2=0.06801...sinx=(0.06801...)(8.2)sinx=0.55773...
    Now we must use the inverse sine operation. This comes with a second group of solutions and a +360°k term. But there is good news: we can ignore the +360°k term. This is because we know the angle must be between zero and 180° because it is in a triangle. However, we cannot ignore the second solution: we must figure out if it gives us a real answer or not. (Remember that the second solution for the sine function comes from 180θ, where θ is whatever angle you get from your calculator.)
    x=sin1(0.55773...)x=33.89952...°or146.10047...°

    We have two answers and we need to figure out which of these answers is correct. Or are they both correct? It turns out that the second answer, x146.1°, is impossible. We know this by comparing to the diagram: the angle we want is an acute angle (less than 90°). So while 146.1° solves the equation, it is not a valid angle for the triangle in this question. (In this case, there is another problem with 146.1°: if we add it to the angle given in the question, we get 146.1°+91°=237.1°. And the angles in a triangle cannot be larger than 180°.)

    NOTE: It is possible that a question like this will have two solutions, or that the obtuse angle is the only correct answer. For example, if the diagram was not drawn to scale, we could not assume that the angle is acute. You should not assume that the answer will always be an acute angle!

    Rounded to the tenths place, the angle has a value of x=33.9°.


    Submit your answer as:

ID is: 1433 Seed is: 6722

The sine rule: identifying the correct sides and angles

The triangle below has sides labelled x,y, and z, and angles labelled X,Y, and Z.

For this triangle, the sine rule states:

?sinX=ysinY

Which of the following choices belongs in the place of the '?' in the formula?

Answer: The missing symbol is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule has an angle and a side. The side must be the side opposite from the angle. Look in the picture of the triangle for sides and angles which are across from (opposite) each other in the triangle.


STEP: Connect the sides and angles in the diagram according to the sine rule
[−1 point ⇒ 0 / 1 points left]

In the sine rule, each fraction must contain a side and an angle. These values (the side and the angle) must be opposite (across from) each other in the triangle. For example, in the triangle in this question, the angle X and side x are opposite each other.

We are given the formula:

?sinX=ysinY

The question mark is in the position of one of the sides. The fraction with the question mark includes the angle X. The side that we need must be opposite from that angle. Looking at the diagram, we can see that the answer is side x.

The missing variable in the equation is the side x.


Submit your answer as:

ID is: 1433 Seed is: 6135

The sine rule: identifying the correct sides and angles

The triangle below has sides labelled c,b, and a, and angles labelled C,B, and A.

For this triangle, the sine rule states:

bsin?=csinC

Which of the following choices belongs in the place of the '?' in the formula?

Answer: The missing symbol is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule has an angle and a side. The side must be the side opposite from the angle. Look in the picture of the triangle for sides and angles which are across from (opposite) each other in the triangle.


STEP: Connect the sides and angles in the diagram according to the sine rule
[−1 point ⇒ 0 / 1 points left]

In the sine rule, each fraction must contain a side and an angle. These values (the side and the angle) must be opposite (across from) each other in the triangle. For example, in the triangle in this question, the angle B and side b are opposite each other.

We are given the formula:

bsin?=csinC

The question mark is in the position of one of the angles. Since the fraction with the question mark has the side b in it, the angle we need must be opposite from side b.

The missing variable in the equation is the angle B.


Submit your answer as:

ID is: 1433 Seed is: 6235

The sine rule: identifying the correct sides and angles

The triangle below has sides labelled a,m, and b, and angles labelled α,θ, and β.

For this triangle, the sine rule states:

msinθ=?sinβ

Which of the following choices belongs in the place of the '?' in the formula?

Answer: The missing symbol is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule has an angle and a side. The side must be the side opposite from the angle. Look in the picture of the triangle for sides and angles which are across from (opposite) each other in the triangle.


STEP: Connect the sides and angles in the diagram according to the sine rule
[−1 point ⇒ 0 / 1 points left]

In the sine rule, each fraction must contain a side and an angle. These values (the side and the angle) must be opposite (across from) each other in the triangle. For example, in the triangle in this question, the angle θ and side m are opposite each other.

We are given the formula:

msinθ=?sinβ

The question mark is in the position of one of the sides. The fraction with the question mark includes the angle β. The side that we need must be opposite from that angle. Looking at the diagram, we can see that the answer is side b.

The missing variable in the equation is the side b.


Submit your answer as:

ID is: 3559 Seed is: 9915

Find the missing length

In triangle EDF, two angles are known: E^=44° and D^=49°. Sides DF¯ and FE¯ have lengths 5w2 and 10w24w respectively, where wR. All lengths are non-zero and have units of cm.

Find then length of DF¯.

INSTRUCTION: Round your answer to two decimal places.
Answer: The length of DF¯ is cm.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by drawing a diagram of the triangle. Label it with everything you know. Then use the sine rule to write an equation which includes the variable w.


STEP: Draw a picture of the triangle and label it
[−1 point ⇒ 4 / 5 points left]

The first thing to do is draw a picture. The question tells us a lot about a triangle, but we need to organise that information in a diagram. It is important to label everything you know about the triangle.

NOTE: The diagram does not need to be to scale. But the labels should be accurate. And since the diagram is not to scale, we cannot draw any conclusions from the picture itself.

STEP: Write an equation including w
[−1 point ⇒ 3 / 5 points left]

We need to determine the value of w. We can do this using the sine rule. We can arrange the sine rule in various ways. In this case, the expression 10w24w can be factorised, and it has a common factor with 5w2! So it looks like we will be able to cancel something. Let's put the side equal to 10w24w in the left numerator so that we can get a more convenient equation to solve.

10w24wsin49°=5w2sin44°

STEP: Solve for w
[−2 points ⇒ 1 / 5 points left]

We need to get the w terms on one side of the equation. After that, we can factorise and simplify on the left side.

10w24wsin49°=5w2sin44°10w24w5w2=sin49°sin44°(2w)(5w2)5w2=sin49°sin44°2w=sin49°sin44°

Now isolate the w on the left side. You can evaluate the sine functions on the right now if you want, but leaving them will be helpful because we can avoid dragging long decimal numbers around.

2w=sin49°sin44°w=(sin49°sin44°)12w=0.54322...

Do not round this number yet - we need to use it in the final calculation.


STEP: Compute the length of DF¯
[−1 point ⇒ 0 / 5 points left]

Finally we can evaluate the length of DF¯.

DF¯=5w2=5(0.54322...)2=0.71611...0.72 cm

As long as this number is positive, it is acceptable (lengths cannot be negative). So we are done.

The length of DF¯ is 0.72 cm.


Submit your answer as:

ID is: 3559 Seed is: 913

Find the missing length

In triangle ACB, two angles are known: A^=78° and C^=70°. Sides CB¯ and BA¯ have lengths 3w+2 and 9w24 respectively, where wR. All lengths are non-zero and have units of cm.

Determine then length of BA¯.

INSTRUCTION: Round your answer to two decimal places.
Answer: The length of BA¯ is cm.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by drawing a diagram of the triangle. Label it with everything you know. Then use the sine rule to write an equation which includes the variable w.


STEP: Draw a picture of the triangle and label it
[−1 point ⇒ 4 / 5 points left]

The first thing to do is draw a picture. The question tells us a lot about a triangle, but we need to organise that information in a diagram. It is important to label everything you know about the triangle.

NOTE: The diagram does not need to be to scale. But the labels should be accurate. And since the diagram is not to scale, we cannot draw any conclusions from the picture itself.

STEP: Write an equation including w
[−1 point ⇒ 3 / 5 points left]

We need to determine the value of w. We can do this using the sine rule. We can arrange the sine rule in various ways. In this case, the expression 9w24 can be factorised, and it has a common factor with 3w+2! So it looks like we will be able to cancel something. Let's put the side equal to 9w24 in the left numerator so that we can get a more convenient equation to solve.

9w24sin70°=3w+2sin78°

STEP: Solve for w
[−2 points ⇒ 1 / 5 points left]

We need to get the w terms on one side of the equation. After that, we can factorise and simplify on the left side.

9w24sin70°=3w+2sin78°9w243w+2=sin70°sin78°(3w2)(3w+2)3w+2=sin70°sin78°3w2=sin70°sin78°

Now isolate the w on the left side. You can evaluate the sine functions on the right now if you want, but leaving them will be helpful because we can avoid dragging long decimal numbers around.

3w2=sin70°sin78°w=(sin70°sin78°+2)13w=0.98689...

Do not round this number yet - we need to use it in the final calculation.


STEP: Compute the length of BA¯
[−1 point ⇒ 0 / 5 points left]

Finally we can evaluate the length of BA¯.

BA¯=9w24=9(0.98689...)24=4.76566...4.77 cm

As long as this number is positive, it is acceptable (lengths cannot be negative). So we are done.

The length of BA¯ is 4.77 cm.


Submit your answer as:

ID is: 3559 Seed is: 6533

Find the missing length

In triangle EDF, two angles are known: E^=69° and D^=50°. Sides DF¯ and FE¯ have lengths 9d24 and 3d+2 respectively, where dR. All lengths are non-zero and have units of mm.

Determine then length of FE¯.

INSTRUCTION: Round your answer to two decimal places.
Answer: The length of FE¯ is mm.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by drawing a diagram of the triangle. Label it with everything you know. Then use the sine rule to write an equation which includes the variable d.


STEP: Draw a picture of the triangle and label it
[−1 point ⇒ 4 / 5 points left]

The first thing to do is draw a picture. The question tells us a lot about a triangle, but we need to organise that information in a diagram. It is important to label everything you know about the triangle.

NOTE: The diagram does not need to be to scale. But the labels should be accurate. And since the diagram is not to scale, we cannot draw any conclusions from the picture itself.

STEP: Write an equation including d
[−1 point ⇒ 3 / 5 points left]

We need to determine the value of d. We can do this using the sine rule. We can arrange the sine rule in various ways. In this case, the expression 9d24 can be factorised, and it has a common factor with 3d+2! So it looks like we will be able to cancel something. Let's put the side equal to 9d24 in the left numerator so that we can get a more convenient equation to solve.

9d24sin69°=3d+2sin50°

STEP: Solve for d
[−2 points ⇒ 1 / 5 points left]

We need to get the d terms on one side of the equation. After that, we can factorise and simplify on the left side.

9d24sin69°=3d+2sin50°9d243d+2=sin69°sin50°(3d2)(3d+2)3d+2=sin69°sin50°3d2=sin69°sin50°

Now isolate the d on the left side. You can evaluate the sine functions on the right now if you want, but leaving them will be helpful because we can avoid dragging long decimal numbers around.

3d2=sin69°sin50°d=(sin69°sin50°+2)13d=1.07290...

Do not round this number yet - we need to use it in the final calculation.


STEP: Compute the length of FE¯
[−1 point ⇒ 0 / 5 points left]

Finally we can evaluate the length of FE¯.

FE¯=3d+2=3(1.07290...)+2=5.21870...5.22 mm

As long as this number is positive, it is acceptable (lengths cannot be negative). So we are done.

The length of FE¯ is 5.22 mm.


Submit your answer as:

ID is: 3581 Seed is: 7381

Connecting the sine rule to right triangles

  1. The sine rule applies to any triangle. This includes right-angled triangles. Which statement below identifies what happens to the sine rule if one of the angles is 90°?

    Choices If an angle is 90°, the sine rule...
    A only works for the acute angles.
    B will give two answers due to the ambiguous case.
    C becomes the same as the theorom of Pythagoras.
    D becomes the sine ratio (opposite & hypotenuse).
    Answer:

    The correct statement is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Draw a triangle to represent the situation described in the question. Then work out the sine rule including the right angle.


    STEP: Find the link between the sine rule and right-angled triangles
    [−1 point ⇒ 0 / 1 points left]

    This question is about what happens when we use a 90° angle in the sine rule. If one of the angles is 90° then we have a right-angled triangle. For example:

    Since this is a right-angled triangle, the normal trigonometric ratios (sine, cosine, and tangent) apply to the triangle. But so does the sine rule. And if we include the right angle when we write the sine rule, it simplifies into the sine ratio for right-angled triangles. (Question 2 is about why this happens.)

    The correct choice is D.


    Submit your answer as:
  2. The table below shows (proves) that if we use the sine rule with a 90° angle it simplifies into the sine ratio for right-angled triangles. However, there is one missing step and one missing reason. Complete the proof by choosing the correct options from the lists available.

    Answer:
    Steps Reasons
    Given
    sinXx=sinZz Sine rule for ΔXYZ
    Substitute Z=90°
    sinXx=1z
    sinX=xz Multiply by x
    sinX=oppositehypotenuse Definition of opposite & hypotenuse
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by analysing the steps and reasons which are already shown. Think about what they mean. Then try to decide what the missing pieces should be.


    STEP: Select the correct step and reason
    [−2 points ⇒ 0 / 2 points left]

    The proof in this question connects the sine rule to right-angled triangles. In such a case, the sine rule becomes the sine ratio for a right-angled triangle. In other words, the sine rule ends up being equivalent to the ratio "opposite over hypotenuse." This connection between the sine rule and the sine ratio exists because sin90° is equal to 1.

    The correct answers in the proof are shown below.

    Steps Reasons
    Given
    sinXx=sinZz Sine rule for ΔXYZ
    sinXx=sin(90°)z Substitute Z=90°
    sinXx=1z Evaluate sin90°=1
    sinX=xz Multiply by x
    sinX=oppositehypotenuse Definition of opposite & hypotenuse

    Submit your answer as: and

ID is: 3581 Seed is: 1269

Connecting the sine rule to right triangles

  1. The sine rule applies to any triangle. This includes right-angled triangles. Which statement below identifies what happens to the sine rule if one of the angles is 90°?

    Choices If an angle is 90°, the sine rule...
    A becomes the sine ratio (opposite & hypotenuse).
    B only works for the acute angles.
    C becomes the same as the theorom of Pythagoras.
    D only uses the reciprocals.
    Answer:

    The correct statement is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Draw a triangle to represent the situation described in the question. Then work out the sine rule including the right angle.


    STEP: Find the link between the sine rule and right-angled triangles
    [−1 point ⇒ 0 / 1 points left]

    This question is about what happens when we use a 90° angle in the sine rule. If one of the angles is 90° then we have a right-angled triangle. For example:

    Since this is a right-angled triangle, the normal trigonometric ratios (sine, cosine, and tangent) apply to the triangle. But so does the sine rule. And if we include the right angle when we write the sine rule, it simplifies into the sine ratio for right-angled triangles. (Question 2 is about why this happens.)

    The correct choice is A.


    Submit your answer as:
  2. The table below shows (proves) that if we use the sine rule with a 90° angle it simplifies into the sine ratio for right-angled triangles. However, there is one missing step and one missing reason. Complete the proof by choosing the correct options from the lists available.

    Answer:
    Steps Reasons
    Given
    sinXx=sinZz
    sinXx=sin(90°)z Substitute Z=90°
    Evaluate sin90°=1
    sinX=xz Multiply by x
    sinX=oppositehypotenuse Definition of opposite & hypotenuse
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by analysing the steps and reasons which are already shown. Think about what they mean. Then try to decide what the missing pieces should be.


    STEP: Select the correct step and reason
    [−2 points ⇒ 0 / 2 points left]

    The proof in this question connects the sine rule to right-angled triangles. In such a case, the sine rule becomes the sine ratio for a right-angled triangle. In other words, the sine rule ends up being equivalent to the ratio "opposite over hypotenuse." This connection between the sine rule and the sine ratio exists because sin90° is equal to 1.

    The correct answers in the proof are shown below.

    Steps Reasons
    Given
    sinXx=sinZz Sine rule for ΔXYZ
    sinXx=sin(90°)z Substitute Z=90°
    sinXx=1z Evaluate sin90°=1
    sinX=xz Multiply by x
    sinX=oppositehypotenuse Definition of opposite & hypotenuse

    Submit your answer as: and

ID is: 3581 Seed is: 6701

Connecting the sine rule to right triangles

  1. The sine rule applies to any triangle. This includes right-angled triangles. Which statement below identifies what happens to the sine rule if one of the angles is 90°?

    Choices If an angle is 90°, the sine rule...
    A becomes the sine ratio (opposite & hypotenuse).
    B will not work anymore.
    C will give two answers due to the ambiguous case.
    D becomes the same as the theorom of Pythagoras.
    Answer:

    The correct statement is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Draw a triangle to represent the situation described in the question. Then work out the sine rule including the right angle.


    STEP: Find the link between the sine rule and right-angled triangles
    [−1 point ⇒ 0 / 1 points left]

    This question is about what happens when we use a 90° angle in the sine rule. If one of the angles is 90° then we have a right-angled triangle. For example:

    Since this is a right-angled triangle, the normal trigonometric ratios (sine, cosine, and tangent) apply to the triangle. But so does the sine rule. And if we include the right angle when we write the sine rule, it simplifies into the sine ratio for right-angled triangles. (Question 2 is about why this happens.)

    The correct choice is A.


    Submit your answer as:
  2. The table below shows (proves) that if we use the sine rule with a 90° angle it simplifies into the sine ratio for right-angled triangles. However, there is one missing step and one missing reason. Complete the proof by choosing the correct options from the lists available.

    Answer:
    Steps Reasons
    Given
    Sine rule for ΔXYZ
    sinXx=sin(90°)z
    sinXx=1z Evaluate sin90°=1
    sinX=xz Multiply by x
    sinX=oppositehypotenuse Definition of opposite & hypotenuse
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by analysing the steps and reasons which are already shown. Think about what they mean. Then try to decide what the missing pieces should be.


    STEP: Select the correct step and reason
    [−2 points ⇒ 0 / 2 points left]

    The proof in this question connects the sine rule to right-angled triangles. In such a case, the sine rule becomes the sine ratio for a right-angled triangle. In other words, the sine rule ends up being equivalent to the ratio "opposite over hypotenuse." This connection between the sine rule and the sine ratio exists because sin90° is equal to 1.

    The correct answers in the proof are shown below.

    Steps Reasons
    Given
    sinXx=sinZz Sine rule for ΔXYZ
    sinXx=sin(90°)z Substitute Z=90°
    sinXx=1z Evaluate sin90°=1
    sinX=xz Multiply by x
    sinX=oppositehypotenuse Definition of opposite & hypotenuse

    Submit your answer as: and

ID is: 3571 Seed is: 8200

The flexibility of the sine rule

The sine rule describes the relationship between the sides and angles in any triangle. The ratio of any side to the sine of the angle opposite from that side is the same for all three side-angle pairs in the triangle. For example, consider this triangle, which has sides m, p, and n, and angles R, S, and Q. Two of the three equal ratios are shown below the triangle.

sinSn=sinQm=third ratio

Which of the choices below is the third equal ratio for this triangle?

A sinRn
B sinRp
C sinSp
Answer: The third ratio is choice .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by figuring out which side and which angle are not used in the two ratios given in the question. They must be in the answer.


STEP: Compare the given ratios to the triangle
[−1 point ⇒ 0 / 1 points left]

This question is about how we can apply the sine rule to a triangle. Each ratio (fraction) in the sine rule contains one side and one angle in the triangle. For any triangle there are three equal ratios. In this question we have two of those ratios for the triangle shown above. We need to find the third equal ratio.

The sine rule always includes side-angle pairs which are across from each other in the triangle. There are always three of these pairs, as shown below:

Each of the linked pairs above corresponds to a ratio for the sine rule. In fact, the two ratios given in the question each include one of the linked pairs:

sinSn=sinQmside n isside m isacross fromacross fromangle Sangle Q

The third side-angle pair is side p and angle R. These values must be in the third ratio!

sinSn=sinQm=sinRp
NOTE:

We can also write the ratios upside down, but if we do this then all the ratios should match:

nsinS=msinQ=psinR

The correct answer is sinRp, which is choice B.


Submit your answer as:

ID is: 3571 Seed is: 1042

The flexibility of the sine rule

The sine rule describes the relationship between the sides and angles in any triangle. The ratio of any side to the sine of the angle opposite from that side is the same for all three side-angle pairs in the triangle. For example, consider this triangle, which has sides p, n, and m, and angles H, G, and F. Two of the three equal ratios are shown below the triangle.

msinG=psinF=third ratio

Which of the choices below is the third equal ratio for this triangle?

A msinH
B nsinF
C nsinH
Answer: The third ratio is choice .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by figuring out which side and which angle are not used in the two ratios given in the question. They must be in the answer.


STEP: Compare the given ratios to the triangle
[−1 point ⇒ 0 / 1 points left]

This question is about how we can apply the sine rule to a triangle. Each ratio (fraction) in the sine rule contains one side and one angle in the triangle. For any triangle there are three equal ratios. In this question we have two of those ratios for the triangle shown above. We need to find the third equal ratio.

The sine rule always includes side-angle pairs which are across from each other in the triangle. There are always three of these pairs, as shown below:

Each of the linked pairs above corresponds to a ratio for the sine rule. In fact, the two ratios given in the question each include one of the linked pairs:

msinG=psinFside m isside p isacross fromacross fromangle Gangle F

The third side-angle pair is side n and angle H. These values must be in the third ratio!

msinG=psinF=nsinH
NOTE:

We can also write the ratios upside down, but if we do this then all the ratios should match:

sinGm=sinFp=sinHn

The correct answer is nsinH, which is choice C.


Submit your answer as:

ID is: 3571 Seed is: 8385

The flexibility of the sine rule

The sine rule describes the relationship between the sides and angles in any triangle. The ratio of any side to the sine of the angle opposite from that side is the same for all three side-angle pairs in the triangle. For example, consider this triangle, which has sides m, p, and n, and angles G, H, and F. Two of the three equal ratios are shown below the triangle.

nsinH=msinF=third ratio

Which of the choices below is the third equal ratio for this triangle?

A psinH
B psinG
C psinF
Answer: The third ratio is choice .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by figuring out which side and which angle are not used in the two ratios given in the question. They must be in the answer.


STEP: Compare the given ratios to the triangle
[−1 point ⇒ 0 / 1 points left]

This question is about how we can apply the sine rule to a triangle. Each ratio (fraction) in the sine rule contains one side and one angle in the triangle. For any triangle there are three equal ratios. In this question we have two of those ratios for the triangle shown above. We need to find the third equal ratio.

The sine rule always includes side-angle pairs which are across from each other in the triangle. There are always three of these pairs, as shown below:

Each of the linked pairs above corresponds to a ratio for the sine rule. In fact, the two ratios given in the question each include one of the linked pairs:

nsinH=msinFside n isside m isacross fromacross fromangle Hangle F

The third side-angle pair is side p and angle G. These values must be in the third ratio!

nsinH=msinF=psinG
NOTE:

We can also write the ratios upside down, but if we do this then all the ratios should match:

sinHn=sinFm=sinGp

The correct answer is psinG, which is choice B.


Submit your answer as:

ID is: 3504 Seed is: 2864

Finding a length with the sine rule

The diagram below shows a triangle with two angles and one side labelled. The given angles are 30° and 52°. The side across from the 30° angle is 5. Determine the length of the side across from the 52° angle. The figure is drawn to scale.

TIP: This diagram is drawn to scale so you can compare the answer you get to the diagram. The side you need is longer than the given side 5. Check your answer before you submit it to make sure it agrees with the figure.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer: The length of the side is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula. Then solve for the missing side.


STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 3 / 4 points left]

The sine rule works for any triangle. It relates four pieces of information in the triangle: two angles and two sides. These angles and sides must be pairs of opposites. We can use the sine rule in either of these ways:

sides in denominators:Angles in numerators andsinA^a=sinB^bangles in denominators:Sides in numerators andasinA^=bsinB^

We should pick the version that makes it easier to solve the question. To find a length, the second version above is the better choice. This is because it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So we will use this arrangement of the formula:

asinA^=bsinB^

STEP: Substitute in the known values
[−1 point ⇒ 2 / 4 points left]

When using the sine rule, we must make sure we substitute the values into the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Each of the arrows above connects numbers that belong together in one fraction. Let's use x to represent the side we are looking for (the one with the '?').

xsin(52°)=5sin(30°)

STEP: Solve for the missing side
[−2 points ⇒ 0 / 4 points left]

Now we can solve for x. First we can evaluate everything on the right side.

xsin(52°)=5sin(30°)xsin(52°)=5(0.5)xsin(52°)=10

Next, we multiply both sides of the equation by sin(52°) to cancel the denominator on the left.

x=(10)sin(52°)x=7.88010...x7.88

The final step rounds the answer to two decimal places, as required in the instructions. Based on the diagram, we expected the answer to be longer than than 5. And it is! Remember to compare your answer to the diagram if the diagram is to scale.

The length of the missing side is 7.88.


Submit your answer as:

ID is: 3504 Seed is: 191

Finding a length with the sine rule

The diagram below shows a triangle with two angles and one side labelled. The given angles are 104° and 29°. The side across from the 104° angle is 8. Determine the length of the side across from the 29° angle. The figure is drawn to scale.

TIP: This diagram is drawn to scale so you can compare the answer you get to the diagram. The side you need is shorter than the given side 8. Check your answer before you submit it to make sure it agrees with the figure.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer: The length of the side is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula. Then solve for the missing side.


STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 3 / 4 points left]

The sine rule works for any triangle. It relates four pieces of information in the triangle: two angles and two sides. These angles and sides must be pairs of opposites. We can use the sine rule in either of these ways:

sides in denominators:Angles in numerators andsinA^a=sinB^bangles in denominators:Sides in numerators andasinA^=bsinB^

We should pick the version that makes it easier to solve the question. To find a length, the second version above is the better choice. This is because it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So we will use this arrangement of the formula:

asinA^=bsinB^

STEP: Substitute in the known values
[−1 point ⇒ 2 / 4 points left]

When using the sine rule, we must make sure we substitute the values into the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Each of the arrows above connects numbers that belong together in one fraction. Let's use x to represent the side we are looking for (the one with the '?').

xsin(29°)=8sin(104°)

STEP: Solve for the missing side
[−2 points ⇒ 0 / 4 points left]

Now we can solve for x. First we can evaluate everything on the right side.

xsin(29°)=8sin(104°)xsin(29°)=8(0.97029...)xsin(29°)=8.24490...

Next, we multiply both sides of the equation by sin(29°) to cancel the denominator on the left.

x=(8.24490...)sin(29°)x=3.99721...x4

The final step rounds the answer to two decimal places, as required in the instructions. Based on the diagram, we expected the answer to be shorter than than 8. And it is! Remember to compare your answer to the diagram if the diagram is to scale.

The length of the missing side is 4.


Submit your answer as:

ID is: 3504 Seed is: 7839

Finding a length with the sine rule

The diagram below shows a triangle with two angles and one side labelled. The given angles are 74° and 44°. The side across from the 44° angle is 8. Determine the length of the side across from the 74° angle. The figure is drawn to scale.

TIP: This diagram is drawn to scale so you can compare the answer you get to the diagram. The side you need is longer than the given side 8. Check your answer before you submit it to make sure it agrees with the figure.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer: The length of the side is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula. Then solve for the missing side.


STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 3 / 4 points left]

The sine rule works for any triangle. It relates four pieces of information in the triangle: two angles and two sides. These angles and sides must be pairs of opposites. We can use the sine rule in either of these ways:

sides in denominators:Angles in numerators andsinA^a=sinB^bangles in denominators:Sides in numerators andasinA^=bsinB^

We should pick the version that makes it easier to solve the question. To find a length, the second version above is the better choice. This is because it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So we will use this arrangement of the formula:

asinA^=bsinB^

STEP: Substitute in the known values
[−1 point ⇒ 2 / 4 points left]

When using the sine rule, we must make sure we substitute the values into the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Each of the arrows above connects numbers that belong together in one fraction. Let's use x to represent the side we are looking for (the one with the '?').

xsin(74°)=8sin(44°)

STEP: Solve for the missing side
[−2 points ⇒ 0 / 4 points left]

Now we can solve for x. First we can evaluate everything on the right side.

xsin(74°)=8sin(44°)xsin(74°)=8(0.69465...)xsin(74°)=11.51645...

Next, we multiply both sides of the equation by sin(74°) to cancel the denominator on the left.

x=(11.51645...)sin(74°)x=11.07032...x11.07

The final step rounds the answer to two decimal places, as required in the instructions. Based on the diagram, we expected the answer to be longer than than 8. And it is! Remember to compare your answer to the diagram if the diagram is to scale.

The length of the missing side is 11.07.


Submit your answer as:

ID is: 3558 Seed is: 6275

Algebra meets the sine rule

In triangle EFD below, two angles are given: E^=61° and D^=70°. Side DE¯ and FD¯ have lengths 3x+9 cm and 7x1 cm respectively, where xR. The third side, EF¯, is 11 cm long. The figure may or may not be drawn to scale.

  1. Determine the value of x.

    INSTRUCTION: Round your answer to two decimal places.
    Answer: x=
    one-of
    type(numeric.abserror(0.004))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by writing down an equation using the sine rule. You should make sure that your equation includes side FD¯, which is equal to 7x1.


    STEP: Write a sine rule equation including x
    [−1 point ⇒ 2 / 3 points left]

    We need to determine the value of x. We can do this by using the sine rule:

    7x1sin(61°)=11sin(70°)

    STEP: Solve the equation
    [−2 points ⇒ 0 / 3 points left]

    Now we can solve the equation. Start by multiplying both sides of the equation by sin(61°). Then work out all the calculations on the right side.

    7x1sin(61°)=11sin(70°)7x1=11sin70°(sin61°)7x1=110.9396...(0.8746...)7x1=10.2382...x=10.2382...+17x=1.6054...1.61

    Rounding to two decimal places, the value of x is 1.61.


    Submit your answer as:
  2. Hence determine the length of DE¯.

    INSTRUCTION:
    • Use the rounded answer from Question 1.
    • Round your answer to two decimal places if necessary.
    Answer: DE¯= cm
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Substitute the answer to Question 1 into the expression for DE¯, which is 3x+9.


    STEP: Substitute and evaluate
    [−1 point ⇒ 0 / 1 points left]

    We need to find the length of side DE¯. It has a length of 3x+9. So we can substitute in the answer from Question 1 and work out the length.

    DE¯=3x+9=3(1.61)+9=13.83 cm

    This value represents a distance (one side of a triangle) so it cannot be negative. Since the answer is positive, everything looks good. (Remember to check your answers if the answer represents a real thing: the answer must be realistic.)

    The length of DE¯ is 13.83 cm.


    Submit your answer as:

ID is: 3558 Seed is: 5463

Algebra meets the sine rule

In triangle ACB below, two angles are given: A^=86° and B^=67°. Side BA¯ and CB¯ have lengths 3y4 cm and 2y4 cm respectively, where yR. The third side, AC¯, is 11 cm long. The figure may or may not be drawn to scale.

  1. Determine the value of y.

    INSTRUCTION: Round your answer to two decimal places.
    Answer: y=
    one-of
    type(numeric.abserror(0.004))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by writing down an equation using the sine rule. You should make sure that your equation includes side CB¯, which is equal to 2y4.


    STEP: Write a sine rule equation including y
    [−1 point ⇒ 2 / 3 points left]

    We need to determine the value of y. We can do this by using the sine rule:

    2y4sin(86°)=11sin(67°)

    STEP: Solve the equation
    [−2 points ⇒ 0 / 3 points left]

    Now we can solve the equation. Start by multiplying both sides of the equation by sin(86°). Then work out all the calculations on the right side.

    2y4sin(86°)=11sin(67°)2y4=11sin67°(sin86°)2y4=110.9205...(0.9975...)2y4=11.9208...y=11.9208...+42y=7.9604...7.96

    Rounding to two decimal places, the value of y is 7.96.


    Submit your answer as:
  2. Hence determine the length of BA¯.

    INSTRUCTION:
    • Use the rounded answer from Question 1.
    • Round your answer to two decimal places if necessary.
    Answer: BA¯= cm
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Substitute the answer to Question 1 into the expression for BA¯, which is 3y4.


    STEP: Substitute and evaluate
    [−1 point ⇒ 0 / 1 points left]

    We need to find the length of side BA¯. It has a length of 3y4. So we can substitute in the answer from Question 1 and work out the length.

    BA¯=3y4=3(7.96)4=19.88 cm

    This value represents a distance (one side of a triangle) so it cannot be negative. Since the answer is positive, everything looks good. (Remember to check your answers if the answer represents a real thing: the answer must be realistic.)

    The length of BA¯ is 19.88 cm.


    Submit your answer as:

ID is: 3558 Seed is: 3544

Algebra meets the sine rule

In triangle FED below, two angles are given: F^=72° and D^=62°. Side DF¯ and ED¯ have lengths 4y7 cm and 3y+3 cm respectively, where yR. The third side, FE¯, is 16 cm long. The figure may or may not be drawn to scale.

  1. What is the value of y?

    INSTRUCTION: Round your answer to two decimal places.
    Answer: y=
    one-of
    type(numeric.abserror(0.004))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by writing down an equation using the sine rule. You should make sure that your equation includes side ED¯, which is equal to 3y+3.


    STEP: Write a sine rule equation including y
    [−1 point ⇒ 2 / 3 points left]

    We need to determine the value of y. We can do this by using the sine rule:

    3y+3sin(72°)=16sin(62°)

    STEP: Solve the equation
    [−2 points ⇒ 0 / 3 points left]

    Now we can solve the equation. Start by multiplying both sides of the equation by sin(72°). Then work out all the calculations on the right side.

    3y+3sin(72°)=16sin(62°)3y+3=16sin62°(sin72°)3y+3=160.8829...(0.9510...)3y+3=17.2342...y=17.2342...33y=4.7447...4.74

    Rounding to two decimal places, the value of y is 4.74.


    Submit your answer as:
  2. Hence determine the length of ED¯.

    INSTRUCTION:
    • Use the rounded answer from Question 1.
    • Round your answer to two decimal places if necessary.
    Answer: ED¯= cm
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Substitute the answer to Question 1 into the expression for ED¯, which is 3y+3.


    STEP: Substitute and evaluate
    [−1 point ⇒ 0 / 1 points left]

    We need to find the length of side ED¯. It has a length of 3y+3. So we can substitute in the answer from Question 1 and work out the length.

    ED¯=3y+3=3(4.74)+3=17.22 cm

    This value represents a distance (one side of a triangle) so it cannot be negative. Since the answer is positive, everything looks good. (Remember to check your answers if the answer represents a real thing: the answer must be realistic.)

    The length of ED¯ is 17.22 cm.


    Submit your answer as:

ID is: 3570 Seed is: 2079

Sine rule problems

In the figure below, WY^X=m, YX^W=n, and YW=d. XY and WZ are perpendicular to ZX.

Determine the distance XY in terms of m, n, and d.

Answer: XY=
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation about triangle YWX. You need this equation to include m, n, and d. So focus on the sides and angles which will bring those variables into the equation.


STEP: Add any information you can to the diagram
[−1 point ⇒ 2 / 3 points left]

We need to find an equation for XY in terms of the three variables m, n, and d. So we need to write an equation about the diagram which includes all four of those things.

How to do that is not obvious because the diagram is a bit complex. Which triangle should we use? Can we take advantage of the right-angles? It is not easy to know at the beginning. So as a first step, we can examine the diagram and label any extra information we can find in the figure.

In this case, we can know four new pieces of information:

  • XY and WZ must be parallel to each other because they are both perpendicular to ZX.
  • ZX^W=90°n because it is complementary to angle n.
  • ZW^X=n because it is an alternate interior angle with n (XY and WZ are parallel).
  • YW^X must be 180°(m+n) because the angles in ΔYWX add up to 180°.

At this point we do not know if any of that information will be useful. We will find out!


STEP: Write an equation and try to include m, n, and d
[−1 point ⇒ 1 / 3 points left]

Now let's write an equation about the diagram. We need an equation which includes the variables m, n, and d (because those must be in the answer). So we need to focus on the sides and angles which will bring those variables into the equation. The equation must also include side XY.

We can try the sine rule. We can get everything we need into the equation if we focus on the sides and angles in ΔYWX.

XYsin(180°(m+n))=dsinn

Now we are in business: the answer must be in terms of m, n. And those are the variables in the equation (in addition to XY).


STEP: Simplify the equation to get the expected form
[−1 point ⇒ 0 / 3 points left]

We need to solve the equation for XY. To do this, multiply both sides of the equation by sin(180°(m+n)).

XY=dsin(180°(m+n))sinn

Now we can use a trigonometric identity to simplify this equation. We know that sin(180°A) is equal to sinA. So the quantity sin(180°(m+n)) is equal to sin(m+n)!

XY=dsin(180°(m+n))sinnXY=dsin(m+n)sinn

That gives us a tidy and accurate expression for XY in terms of the required values.

The length of XY, as a function of m, n, and d is dsin(m+n)sinn.


Submit your answer as:

ID is: 3570 Seed is: 7555

Sine rule problems

In the figure below, ZX^W=f, XW^Z=g, and XZ=d. WX and ZY are perpendicular to YW.

Determine the distance WX in terms of f, g, and d.

Answer: WX=
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation about triangle XZW. You need this equation to include f, g, and d. So focus on the sides and angles which will bring those variables into the equation.


STEP: Add any information you can to the diagram
[−1 point ⇒ 2 / 3 points left]

We need to find an equation for WX in terms of the three variables f, g, and d. So we need to write an equation about the diagram which includes all four of those things.

How to do that is not obvious because the diagram is a bit complex. Which triangle should we use? Can we take advantage of the right-angles? It is not easy to know at the beginning. So as a first step, we can examine the diagram and label any extra information we can find in the figure.

In this case, we can know four new pieces of information:

  • WX and ZY must be parallel to each other because they are both perpendicular to YW.
  • YW^Z=90°g because it is complementary to angle g.
  • YZ^W=g because it is an alternate interior angle with g (WX and ZY are parallel).
  • XZ^W must be 180°(f+g) because the angles in ΔXZW add up to 180°.

At this point we do not know if any of that information will be useful. We will find out!


STEP: Write an equation and try to include f, g, and d
[−1 point ⇒ 1 / 3 points left]

Now let's write an equation about the diagram. We need an equation which includes the variables f, g, and d (because those must be in the answer). So we need to focus on the sides and angles which will bring those variables into the equation. The equation must also include side WX.

We can try the sine rule. We can get everything we need into the equation if we focus on the sides and angles in ΔXZW.

WXsin(180°(f+g))=dsing

Now we are in business: the answer must be in terms of f, g. And those are the variables in the equation (in addition to WX).


STEP: Simplify the equation to get the expected form
[−1 point ⇒ 0 / 3 points left]

We need to solve the equation for WX. To do this, multiply both sides of the equation by sin(180°(f+g)).

WX=dsin(180°(f+g))sing

Now we can use a trigonometric identity to simplify this equation. We know that sin(180°A) is equal to sinA. So the quantity sin(180°(f+g)) is equal to sin(f+g)!

WX=dsin(180°(f+g))singWX=dsin(f+g)sing

That gives us a tidy and accurate expression for WX in terms of the required values.

The length of WX, as a function of f, g, and d is dsin(f+g)sing.


Submit your answer as:

ID is: 3570 Seed is: 4720

Sine rule problems

In the figure below, RP^Q=f, PQ^R=g, and QP=h. QP and RS are perpendicular to SQ.

Determine the distance RQ in terms of f, g, and h.

Answer: RQ=
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation about triangle PRQ. You need this equation to include f, g, and h. So focus on the sides and angles which will bring those variables into the equation.


STEP: Add any information you can to the diagram
[−1 point ⇒ 2 / 3 points left]

We need to find an equation for RQ in terms of the three variables f, g, and h. So we need to write an equation about the diagram which includes all four of those things.

How to do that is not obvious because the diagram is a bit complex. Which triangle should we use? Can we take advantage of the right-angles? It is not easy to know at the beginning. So as a first step, we can examine the diagram and label any extra information we can find in the figure.

In this case, we can know four new pieces of information:

  • QP and RS must be parallel to each other because they are both perpendicular to SQ.
  • SQ^R=90°g because it is complementary to angle g.
  • SR^Q=g because it is an alternate interior angle with g (QP and RS are parallel).
  • PR^Q must be 180°(f+g) because the angles in ΔPRQ add up to 180°.

At this point we do not know if any of that information will be useful. We will find out!


STEP: Write an equation and try to include f, g, and h
[−1 point ⇒ 1 / 3 points left]

Now let's write an equation about the diagram. We need an equation which includes the variables f, g, and h (because those must be in the answer). So we need to focus on the sides and angles which will bring those variables into the equation. The equation must also include side RQ.

We can try the sine rule. We can get everything we need into the equation if we focus on the sides and angles in ΔPRQ.

RQsinf=hsin(180°(f+g))

Now we are in business: the answer must be in terms of f, g. And those are the variables in the equation (in addition to RQ).


STEP: Simplify the equation to get the expected form
[−1 point ⇒ 0 / 3 points left]

We need to solve the equation for RQ. To do this, multiply both sides of the equation by sinf.

RQ=hsinfsin(180°(f+g))

Now we can use a trigonometric identity to simplify this equation. We know that sin(180°A) is equal to sinA. So the quantity sin(180°(f+g)) is equal to sin(f+g)!

RQ=hsinfsin(180°(f+g))RQ=hsinfsin(f+g)

That gives us a tidy and accurate expression for RQ in terms of the required values.

The length of RQ, as a function of f, g, and h is hsinfsin(f+g).


Submit your answer as:

2. Cosine rule


ID is: 3584 Seed is: 4587

Perimeter problems

Triangle SQR has one angle and two sides known: Q^=96.1° is the angle between the sides with lengths 11.3 mm and 19.4 mm. The figure is not drawn to scale.

Find the perimeter of the triangle. Then select the correct units to go with the answer.

INSTRUCTION: Round your answer to two decimal places.
Answer:

The perimeter is .

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

To find the perimeter you need to find the third side of the triangle.


STEP: Use the cosine rule to find the missing side of the triangle
[−2 points ⇒ 2 / 4 points left]

The perimeter of a shape is the distance around the shape. To find the perimeter of the triangle we must find the length of the third side. We can do that with the cosine rule. Let q represent the length of the side we want (the side across from Q^).

q2=(11.3)2+(19.4)22(11.3)(19.4)cos96.1°q2=127.69+376.36438.44cos96.1°q2=679.94531.1cos96.1°q2=550.64041...q2=±550.64041...q=±23.46572...

Do not round off yet: we always want to wait to get the final answer before rounding off. However, we can drop the negative answer, because the distance must be positive.


STEP: Calculate the perimeter
[−1 point ⇒ 1 / 4 points left]

The perimeter is the sum of all three sides:

Perimeter =11.3+19.4+23.46572...=54.16572...54.17

STEP: Select the correct units
[−1 point ⇒ 0 / 4 points left]

The only thing left to do is to pick the correct units for the perimeter. Perimeter is a distance, so the units will always be linear units (not squared or cubed). In this case, the units are mm.

The perimeter of the triangle is 54.17 mm.


Submit your answer as: and

ID is: 3584 Seed is: 6640

Perimeter problems

Triangle LMK has one angle and two sides known: M^=109.4° is the angle between the sides with lengths 12.7 cm and 20.3 cm. The figure is not drawn to scale.

Calculate the perimeter of the triangle. Then select the correct units to go with the answer.

INSTRUCTION: Round your answer to two decimal places.
Answer:

The perimeter is .

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

To find the perimeter you need to find the third side of the triangle.


STEP: Use the cosine rule to find the missing side of the triangle
[−2 points ⇒ 2 / 4 points left]

The perimeter of a shape is the distance around the shape. To find the perimeter of the triangle we must find the length of the third side. We can do that with the cosine rule. Let m represent the length of the side we want (the side across from M^).

m2=(12.7)2+(20.3)22(12.7)(20.3)cos109.4°m2=161.29+412.09515.62cos109.4°m2=906.58693.42cos109.4°m2=744.64892...m2=±744.64892...m=±27.28825...

Do not round off yet: we always want to wait to get the final answer before rounding off. However, we can drop the negative answer, because the distance must be positive.


STEP: Calculate the perimeter
[−1 point ⇒ 1 / 4 points left]

The perimeter is the sum of all three sides:

Perimeter =12.7+20.3+27.28825...=60.28825...60.29

STEP: Select the correct units
[−1 point ⇒ 0 / 4 points left]

The only thing left to do is to pick the correct units for the perimeter. Perimeter is a distance, so the units will always be linear units (not squared or cubed). In this case, the units are cm.

The perimeter of the triangle is 60.29 cm.


Submit your answer as: and

ID is: 3584 Seed is: 8250

Perimeter problems

Triangle RSQ has one angle and two sides known: S^=109.7° is the angle between the sides with lengths 15.2 mm and 21 mm. The figure is not drawn to scale.

Find the perimeter of the triangle. Then select the correct units to go with the answer.

INSTRUCTION: Round your answer to two decimal places.
Answer:

The perimeter is .

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

To find the perimeter you need to find the third side of the triangle.


STEP: Use the cosine rule to find the missing side of the triangle
[−2 points ⇒ 2 / 4 points left]

The perimeter of a shape is the distance around the shape. To find the perimeter of the triangle we must find the length of the third side. We can do that with the cosine rule. Let s represent the length of the side we want (the side across from S^).

s2=(15.2)2+(21)22(15.2)(21)cos109.7°s2=231.04+441638.4cos109.7°s2=1,119.08905.92cos109.7°s2=887.24161...s2=±887.24161...s=±29.78660...

Do not round off yet: we always want to wait to get the final answer before rounding off. However, we can drop the negative answer, because the distance must be positive.


STEP: Calculate the perimeter
[−1 point ⇒ 1 / 4 points left]

The perimeter is the sum of all three sides:

Perimeter =15.2+21+29.78660...=65.98660...65.99

STEP: Select the correct units
[−1 point ⇒ 0 / 4 points left]

The only thing left to do is to pick the correct units for the perimeter. Perimeter is a distance, so the units will always be linear units (not squared or cubed). In this case, the units are mm.

The perimeter of the triangle is 65.99 mm.


Submit your answer as: and

ID is: 4394 Seed is: 4329

Application of cos rule to prove an identity

Adapted from DBE Nov 2016, Grade 11, P2, Q7
Maths formulas

Given the cos rule, c2=a2+b22abcosC, Bianca was asked to prove that:

1+cosC = (a+b+c)(a+bc)2ab

Bianca has already answered the question below. But, she has made a mistake. Here is Bianca's answer, labelled Line 1 to Line 8:

1:c2=a2+b22abcosC2:2abcosC=a2+b2c23:cosC=a2+b2c22ab4:1+cosC=1+a2+b2c22ab5:1+cosC=2ab+a2+b2c22ab6:1+cosC=[(a2+2ab+b2)(c)2]2ab7:1+cosC=[(a+b)2(c)2][÷2ab]8:1+cosC=(a+b+c)(a+bc)2ab
Answer:
  1. The first line containing an error is .
  2. The correct replacement for this line is:
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write out the cos rule and use it to find an expression for 1+cosC.


STEP: Write out the cos rule and use it to find an expression for 1+cosC
[−4 points ⇒ 0 / 4 points left]

The incorrect proof is:

1:c2=a2+b22abcosC2:2abcosC=a2+b2c23:cosC=a2+b2c22ab4:1+cosC=1+a2+b2c22ab5:1+cosC=2ab+a2+b2c22ab6:1+cosC=[(a2+2ab+b2)(c)2]2ab7:1+cosC=[(a+b)2(c)2][÷2ab]8:1+cosC=(a+b+c)(a+bc)2ab

The first line that contains an error is Line 3.

When making cosC the subject of the expression, we should divide by positive 2ab, not subtract 2ab.

The correct Line 3 is:

cosC=a2+b2c22ab

Here are the steps explained:

  • Write out the cos rule.
  • Make cosC the subject.
  • Add 1 to both sides of the equation.
  • Group RHS terms to identify a difference of two squares.
  • Factorise RHS using difference of squares to complete the proof.

The correct proof is:

1:c2=a2+b22abcosC2:2abcosC=a2+b2c23:cosC=a2+b2c22ab4:1+cosC=1+a2+b2c22ab5:1+cosC=2ab+a2+b2c22ab6:1+cosC=(a2+2ab+b2)(c)22ab7:1+cosC=(a+b)2(c)22ab8:1+cosC=(a+b+c)(a+bc)2ab
TIP: Spotting a "difference of two squares" and factorising it correctly can be tricky. You need to have two squared-terms separated by a negative sign, like a2b2. When factorised, this becomes (a+b)(ab).

You can use more brackets for the tricky ones, to help you to remember that "1×(bracket)" means multiply all terms in the bracket by 1.

For example:

x2(y+3)2=[x+(y+3)][x(y+3)]=(x+y+3)(xy3)

Submit your answer as: and

ID is: 4394 Seed is: 9262

Application of cos rule to prove an identity

Adapted from DBE Nov 2016, Grade 11, P2, Q7
Maths formulas

Given the cos rule, c2=b2+a22bacosC, Megan was asked to prove that:

1+cosC = (b+a+c)(b+ac)2ba

Megan has already answered the question below. But, she has made a mistake. Here is Megan's answer, labelled Line 1 to Line 8:

1:c2=b2+a22bacosC2:2bacosC=b2+a2c23:cosC=b2+a2c22ba4:1+cosC=1+b2+a2c22ba5:1+cosC=1+b2+a2c22ba6:1+cosC=(b2+1+a2)(c)22ba7:1+cosC=(b+1+a)2(c)22ba8:1+cosC=(b+a+c)(b+ac)2ba
Answer:
  1. The first line containing an error is .
  2. The correct replacement for this line is:
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write out the cos rule and use it to find an expression for 1+cosC.


STEP: Write out the cos rule and use it to find an expression for 1+cosC
[−4 points ⇒ 0 / 4 points left]

The incorrect proof is:

1:c2=b2+a22bacosC2:2bacosC=b2+a2c23:cosC=b2+a2c22ba4:1+cosC=1+b2+a2c22ba5:1+cosC=1+b2+a2c22ba6:1+cosC=(b2+1+a2)(c)22ba7:1+cosC=(b+1+a)2(c)22ba8:1+cosC=(b+a+c)(b+ac)2ba

The first line that contains an error is Line 5.

When adding 1 to a fraction it is incorrect to just add 1 to the numerator, we should add 2ba2ba.

The correct Line 5 is:

1+cosC=2ba+b2+a2c22ba

Here are the steps explained:

  • Write out the cos rule.
  • Make cosC the subject.
  • Add 1 to both sides of the equation.
  • Group RHS terms to identify a difference of two squares.
  • Factorise RHS using difference of squares to complete the proof.

The correct proof is:

1:c2=b2+a22bacosC2:2bacosC=b2+a2c23:cosC=b2+a2c22ba4:1+cosC=1+b2+a2c22ba5:1+cosC=2ba+b2+a2c22ba6:1+cosC=(b2+2ba+a2)(c)22ba7:1+cosC=(b+a)2(c)22ba8:1+cosC=(b+a+c)(b+ac)2ba
TIP: Spotting a "difference of two squares" and factorising it correctly can be tricky. You need to have two squared-terms separated by a negative sign, like b2a2. When factorised, this becomes (b+a)(ba).

You can use more brackets for the tricky ones, to help you to remember that "1×(bracket)" means multiply all terms in the bracket by 1.

For example:

x2(y+3)2=[x+(y+3)][x(y+3)]=(x+y+3)(xy3)

Submit your answer as: and

ID is: 4394 Seed is: 7097

Application of cos rule to prove an identity

Adapted from DBE Nov 2016, Grade 11, P2, Q7
Maths formulas

Given the cos rule, b2=a2+c22accosB, Lethabo was asked to prove that:

1cosB = (ba+c)(b+ac)2ac

Lethabo has already answered the question below. But, he has made a mistake. Here is Lethabo's answer, labelled Line 1 to Line 8:

1:b2=a2+c22accosB2:2accosB=a2+c2b23:cosB=a2+c2b22ac4:1cosB=1a2+c2b22ac5:1cosB=2aca2c2+b22ac6:1cosB=a22ac+c2b22ac7:1cosB=(ac)2b22ac8:1cosB=(acb)(ac+b)2ac
Answer:
  1. The first line containing an error is .
  2. The correct replacement for this line is:
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Write out the cos rule and use it to find an expression for 1cosB.


STEP: Write out the cos rule and use it to find an expression for 1cosB
[−4 points ⇒ 0 / 4 points left]

The incorrect proof is:

1:b2=a2+c22accosB2:2accosB=a2+c2b23:cosB=a2+c2b22ac4:1cosB=1a2+c2b22ac5:1cosB=2aca2c2+b22ac6:1cosB=a22ac+c2b22ac7:1cosB=(ac)2b22ac8:1cosB=(acb)(ac+b)2ac

The first line that contains an error is Line 6.

Line 5 is correct but in Line 6 there are errors with the signs for example, the a2 term suddenly changed to a2!

The correct Line 6 is:

1cosB=b2(a22ac+c2)2ac

Here are the steps explained:

  • Write out the cos rule.
  • Make cosB the subject.
  • Begin both sides with 1.
  • Group RHS terms to identify a difference of two squares.
  • Factorise RHS to complete the proof.

The correct proof is:

1:b2=a2+c22accosB2:2accosB=a2+c2b23:cosB=a2+c2b22ac4:1cosB=1a2+c2b22ac5:1cosB=2aca2c2+b22ac6:1cosB=b2(a22ac+c2)2ac7:1cosB=b2(ac)22ac8:1cosB=(b(ac))(b+(ac))2ac9:1cosB=(ba+c)(b+ac)2ac
TIP: Spotting a "difference of two squares" and factorising it correctly can be tricky. You need to have two squared-terms separated by a negative sign, like a2c2. When factorised, this becomes (a+c)(ac).

You can use more brackets for the tricky ones, to help you to remember that "1×(bracket)" means multiply all terms in the bracket by 1.

For example:

x2(y+3)2=[x+(y+3)][x(y+3)]=(x+y+3)(xy3)

Submit your answer as: and

ID is: 3589 Seed is: 3834

Figuring out what information to use

The figure below shows triangle ACB and triangle ADB, which have a common side AB¯. In triangle ACB, we have the sides BC¯=3.7, CA¯=5.1, and AB¯=4.7. In triangle ADB we know side DB¯=3.6 and angle BAD=47°. CAB is unknown and labelled '?'.

Answer the two questions below about this diagram and the cosine rule.

Answer:
  1. Can you find the missing angle with the cosine rule?
  2. How do you know? .
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The cosine rule includes three sides and one angle in a triangle. If you know any three of those values, you can find the fourth value. Do you have three of those four pieces of information in this question?


STEP: Compare the information given to what the cosine rule requires
[−1 point ⇒ 1 / 2 points left]

We need to determine if we can calculate the missing angle using the cosine rule. The missing angle is part of triangle ACB. So we can ignore the other triangle.

Now compare the information in this picture to the cosine rule: the cosine rule includes all three sides and one angle in the triangle. If we know three of those values, we can calculate the fourth value. But if we know fewer than three of those values we do not have enough information to use the cosine rule.

In this question, we know three of the four values in the cosine rule (the values of a, b, and c). So we can use the cosine rule to calculate the value of the unknown angle.


STEP: Select the correct answers
[−1 point ⇒ 0 / 2 points left]

Based on the reasons above we can select the correct answer choices:

  1. Can you find the missing angle with the cosine rule? Yes
  2. We know this because we know 3 of the 4 values in the cosine rule.

Submit your answer as: and

ID is: 3589 Seed is: 2830

Figuring out what information to use

The figure below shows triangle CBA and triangle CDA, which have a common side CA¯. In triangle CDA we know side CD¯=4.4 and angle CDA=65°. In triangle CBA, we know sides CA¯=4.7 and BC¯=5.6, and we know angles CBA=54° and BCA=43°. AB¯ is unknown and labelled '?'.

Answer the two questions below about this diagram and the cosine rule.

Answer:
  1. Can you find the missing side with the cosine rule?
  2. How do you know? .
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The cosine rule includes three sides and one angle in a triangle. If you know any three of those values, you can find the fourth value. Do you have three of those four pieces of information in this question?


STEP: Compare the information given to what the cosine rule requires
[−1 point ⇒ 1 / 2 points left]

We need to determine if we can calculate the missing side using the cosine rule. The missing side is part of triangle CBA. So we can ignore the other triangle.

Now compare the information in this picture to the cosine rule: the cosine rule includes all three sides and one angle in the triangle. If we know three of those values, we can calculate the fourth value. But if we know fewer than three of those values we do not have enough information to use the cosine rule.

In this question, we know three of the four values in the cosine rule (the sides a and b, as well as the angle θ) So we can use the cosine rule to calculate the value of the unknown side.

NOTE: The second angle in the triangle (the 43° angle) is extra information: we do not need it in the cosine rule.

STEP: Select the correct answers
[−1 point ⇒ 0 / 2 points left]

Based on the reasons above we can select the correct answer choices:

  1. Can you find the missing side with the cosine rule? Yes
  2. We know this because we know 3 of the 4 values in the cosine rule.

Submit your answer as: and

ID is: 3589 Seed is: 1989

Figuring out what information to use

The figure below shows triangle CBD and triangle CAD, which have a common side CD¯. In triangle CAD we know side CA¯=3.6 and angle DCA=37°. In triangle CBD, we have the sides DB¯=3.9, BC¯=3.5, and CD¯=4.7. BCD is unknown and labelled '?'.

Answer the two questions below about this diagram and the cosine rule.

Answer:
  1. Can you find the missing angle with the cosine rule?
  2. How do you know? .
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The cosine rule includes three sides and one angle in a triangle. If you know any three of those values, you can find the fourth value. Do you have three of those four pieces of information in this question?


STEP: Compare the information given to what the cosine rule requires
[−1 point ⇒ 1 / 2 points left]

We need to determine if we can calculate the missing angle using the cosine rule. The missing angle is part of triangle CBD. So we can ignore the other triangle.

Now compare the information in this picture to the cosine rule: the cosine rule includes all three sides and one angle in the triangle. If we know three of those values, we can calculate the fourth value. But if we know fewer than three of those values we do not have enough information to use the cosine rule.

In this question, we know three of the four values in the cosine rule (the values of a, b, and c). So we can use the cosine rule to calculate the value of the unknown angle.


STEP: Select the correct answers
[−1 point ⇒ 0 / 2 points left]

Based on the reasons above we can select the correct answer choices:

  1. Can you find the missing angle with the cosine rule? Yes
  2. We know this because we know 3 of the 4 values in the cosine rule.

Submit your answer as: and

ID is: 3505 Seed is: 4151

Finding the side opposite a known angle

The triangle below has two sides given: AC¯=6 and BC¯=8. The angle between these two sides is C^=58°. These values are labelled in the diagram. The figure is drawn to scale.

Compute the length of the third side of the triangle, BA¯.

TIP: The question states that the figure is drawn to scale. The third side looks like it is a similar size to the other sides. Therefore your answer should be a number somehow close to 6 or 8. If your answer does not agree with this, it probably means there is a mistake in your work.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer:

The length of side BA¯ is units.

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The cosine rule is:

c2=a2+b22abcosθ

Start by figuring out where to substitute in the two sides and the angle given in the question.


STEP: Identify the side opposite the angle
[−1 point ⇒ 3 / 4 points left]

We can answer this question using the cosine rule. This is because the cosine rule relates all three sides of the triangle and one of the angles.

The cosine rule says:

c2=a2+b22abcosθ

where θ is one of the angles in the triangle and c represents the side of the triangle opposite to that angle. When we use the cosine rule, it is always important to identify the side of the triangle which is opposite from the angle you are interested in.

In this question, the only angle given is C^=58°. So the side we need to identify is across from the angle labelled as 58°. The picture below shows that this side is across the triangle from (opposite to) the angle.


STEP: Substitute into the formula
[−1 point ⇒ 2 / 4 points left]

Based on the picture above, we can substitute the values into the formula. Here are the values that we know:

  • The angle θ is 58°.
  • c is the side labelled with the "?". We will use x for this side (you can use any variable you want).
  • a and b are the other sides of the triangle. It does not matter which side we use for a and which for b.

Substituting into the formula, we get:

c2=a2+b22abcosθx2=(6)2+(8)22(6)(8)cos(58°)

STEP: Substitute into the formula
[−2 points ⇒ 0 / 4 points left]

Now we need to solve for the value of x. Start by evaluating all the calculations on the right side. Do not round off during these calcations.

x2=(6)2+(8)22(6)(8)cos(58°)x2=36+64(96)(0.52991...)x2=10050.87224...x2=49.12775...

Now we must square root both sides of the equation to change x2 to x. When we do this, we must remember to bring in a plus-minus (this is required any time we square root when solving an equation).

x2=49.12775...x=±49.12775...x=±7.00911...Which means:x=7.00911...or7.00911...

Finally we have the answer. But there are two different answers for this question: 7.00911... and 7.00911...! Are they both correct? Remember that the answer represents the side of a triangle. And a length cannot be negative. So we must throw out the negative solution.

The last thing to do is round off the answer to two decimal places (as required by the instructions). This gives us: 7.00911...x=7.01.

Therefore, the correct answer, rounded to two places, is: x=7.01 units.


Submit your answer as:

ID is: 3505 Seed is: 3779

Finding the side opposite a known angle

The triangle below has two sides given: QP¯=6 and QR¯=9. The angle between these two sides is Q^=51°. These values are labelled in the diagram. The figure is drawn to scale.

Calculate the length of the third side of the triangle, RP¯.

TIP: The question states that the figure is drawn to scale. The third side is similar in length to the side with length 6. Therefore your answer should be somewhat close to 6. If your answer does not agree with this, it probably means there is a mistake in your work.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer:

The length of side RP¯ is units.

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The cosine rule is:

c2=a2+b22abcosθ

Start by figuring out where to substitute in the two sides and the angle given in the question.


STEP: Identify the side opposite the angle
[−1 point ⇒ 3 / 4 points left]

We can answer this question using the cosine rule. This is because the cosine rule relates all three sides of the triangle and one of the angles.

The cosine rule says:

c2=a2+b22abcosθ

where θ is one of the angles in the triangle and c represents the side of the triangle opposite to that angle. When we use the cosine rule, it is always important to identify the side of the triangle which is opposite from the angle you are interested in.

In this question, the only angle given is Q^=51°. So the side we need to identify is across from the angle labelled as 51°. The picture below shows that this side is across the triangle from (opposite to) the angle.


STEP: Substitute into the formula
[−1 point ⇒ 2 / 4 points left]

Based on the picture above, we can substitute the values into the formula. Here are the values that we know:

  • The angle θ is 51°.
  • c is the side labelled with the "?". We will use x for this side (you can use any variable you want).
  • a and b are the other sides of the triangle. It does not matter which side we use for a and which for b.

Substituting into the formula, we get:

c2=a2+b22abcosθx2=(6)2+(9)22(6)(9)cos(51°)

STEP: Substitute into the formula
[−2 points ⇒ 0 / 4 points left]

Now we need to solve for the value of x. Start by evaluating all the calculations on the right side. Do not round off during these calcations.

x2=(6)2+(9)22(6)(9)cos(51°)x2=36+81(108)(0.62932...)x2=11767.96660...x2=49.03339...

Now we must square root both sides of the equation to change x2 to x. When we do this, we must remember to bring in a plus-minus (this is required any time we square root when solving an equation).

x2=49.03339...x=±49.03339...x=±7.00238...Which means:x=7.00238...or7.00238...

Finally we have the answer. But there are two different answers for this question: 7.00238... and 7.00238...! Are they both correct? Remember that the answer represents the side of a triangle. And a length cannot be negative. So we must throw out the negative solution.

The last thing to do is round off the answer to two decimal places (as required by the instructions). This gives us: 7.00238...x=7.

Therefore, the correct answer, rounded to two places, is: x=7 units.


Submit your answer as:

ID is: 3505 Seed is: 6853

Finding the side opposite a known angle

The triangle below has two sides given: YZ¯=9 and ZX¯=10. The angle between these two sides is Z^=43°. These values are labelled in the diagram. The figure is drawn to scale.

Determine the length of the third side of the triangle, YX¯.

TIP: The question states that the figure is drawn to scale. The third side is clearly shorter than either of the given sides. So your answer should be less than 9. If your answer does not agree with this, it probably means there is a mistake in your work.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer:

The length of side YX¯ is units.

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The cosine rule is:

c2=a2+b22abcosθ

Start by figuring out where to substitute in the two sides and the angle given in the question.


STEP: Identify the side opposite the angle
[−1 point ⇒ 3 / 4 points left]

We can answer this question using the cosine rule. This is because the cosine rule relates all three sides of the triangle and one of the angles.

The cosine rule says:

c2=a2+b22abcosθ

where θ is one of the angles in the triangle and c represents the side of the triangle opposite to that angle. When we use the cosine rule, it is always important to identify the side of the triangle which is opposite from the angle you are interested in.

In this question, the only angle given is Z^=43°. So the side we need to identify is across from the angle labelled as 43°. The picture below shows that this side is across the triangle from (opposite to) the angle.


STEP: Substitute into the formula
[−1 point ⇒ 2 / 4 points left]

Based on the picture above, we can substitute the values into the formula. Here are the values that we know:

  • The angle θ is 43°.
  • c is the side labelled with the "?". We will use x for this side (you can use any variable you want).
  • a and b are the other sides of the triangle. It does not matter which side we use for a and which for b.

Substituting into the formula, we get:

c2=a2+b22abcosθx2=(9)2+(10)22(9)(10)cos(43°)

STEP: Substitute into the formula
[−2 points ⇒ 0 / 4 points left]

Now we need to solve for the value of x. Start by evaluating all the calculations on the right side. Do not round off during these calcations.

x2=(9)2+(10)22(9)(10)cos(43°)x2=81+100(180)(0.73135...)x2=181131.64366...x2=49.35633...

Now we must square root both sides of the equation to change x2 to x. When we do this, we must remember to bring in a plus-minus (this is required any time we square root when solving an equation).

x2=49.35633...x=±49.35633...x=±7.02540...Which means:x=7.02540...or7.02540...

Finally we have the answer. But there are two different answers for this question: 7.02540... and 7.02540...! Are they both correct? Remember that the answer represents the side of a triangle. And a length cannot be negative. So we must throw out the negative solution.

The last thing to do is round off the answer to two decimal places (as required by the instructions). This gives us: 7.02540...x=7.03.

Therefore, the correct answer, rounded to two places, is: x=7.03 units.


Submit your answer as:

ID is: 3552 Seed is: 6086

The cosine rule: finding mistakes

The triangle below has sides 4, 5, and 6, and one angle labelled as C. A friend of yours writes down the cosine rule for this triangle, but there is a mistake.

Incorrect equation:

42=62+522(6)(5)cosC

How can your friend correct the equation?

Answer:

To fix the equation, your friend should .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

For the cosine rule, the side of the triangle opposite the angle must go into the equation on the opposite side of the angle. Which side of the triangle is opposite from angle C?


STEP: Identify the side opposite from the given angle
[−1 point ⇒ 0 / 1 points left]

The cosine rule is an equation about triangles. It includes all three sides and one angle:

c2=a2+b22abcosθ

where:

  • θ is an angle in the triangle
  • c is the side of the triangle opposite from θ
  • a and b are the sides adjacent to θ

For the cosine rule, one side is more important that the other two:

The side opposite from the angle in the triangle must be on the opposite side of the equation. And that is why the equation in the question is wrong:

42=6_2+522(6_)(5)cosC_

The higlighted side and angle should not be on the same side of the equation, since the side 6 is opposite the angle C.

So 6 must go in the correct spot. But sides 4 and 5 are more flexible. The correct equation for this triangle can be either of these:

62=42+522(4)(5)cosCor62=52+422(5)(4)cosC

To fix the equation, swap 6 with 4.


Submit your answer as:

ID is: 3552 Seed is: 3409

The cosine rule: finding mistakes

The triangle below has sides 6, 7, and 10, and one angle labelled as C. A friend of yours writes down the cosine rule for this triangle, but there is a mistake.

Incorrect equation:

62=72+1022(7)(10)cosC

How can your friend correct the equation?

Answer:

To fix the equation, your friend should .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

For the cosine rule, the side of the triangle opposite the angle must go into the equation on the opposite side of the angle. Which side of the triangle is opposite from angle C?


STEP: Identify the side opposite from the given angle
[−1 point ⇒ 0 / 1 points left]

The cosine rule is an equation about triangles. It includes all three sides and one angle:

c2=a2+b22abcosθ

where:

  • θ is an angle in the triangle
  • c is the side of the triangle opposite from θ
  • a and b are the sides adjacent to θ

For the cosine rule, one side is more important that the other two:

The side opposite from the angle in the triangle must be on the opposite side of the equation. And that is why the equation in the question is wrong:

62=7_2+1022(7_)(10)cosC_

The higlighted side and angle should not be on the same side of the equation, since the side 7 is opposite the angle C.

So 7 must go in the correct spot. But sides 6 and 10 are more flexible. The correct equation for this triangle can be either of these:

72=62+1022(6)(10)cosCor72=102+622(10)(6)cosC

To fix the equation, swap 7 with 6.


Submit your answer as:

ID is: 3552 Seed is: 8832

The cosine rule: finding mistakes

The triangle below has sides 5, 6, and 7, and one angle labelled as C. A friend of yours writes down the cosine rule for this triangle, but there is a mistake.

Incorrect equation:

62=52+722(5)(7)cosC

How can your friend correct the equation?

Answer:

To fix the equation, your friend should .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

For the cosine rule, the side of the triangle opposite the angle must go into the equation on the opposite side of the angle. Which side of the triangle is opposite from angle C?


STEP: Identify the side opposite from the given angle
[−1 point ⇒ 0 / 1 points left]

The cosine rule is an equation about triangles. It includes all three sides and one angle:

c2=a2+b22abcosθ

where:

  • θ is an angle in the triangle
  • c is the side of the triangle opposite from θ
  • a and b are the sides adjacent to θ

For the cosine rule, one side is more important that the other two:

The side opposite from the angle in the triangle must be on the opposite side of the equation. And that is why the equation in the question is wrong:

62=5_2+722(5_)(7)cosC_

The higlighted side and angle should not be on the same side of the equation, since the side 5 is opposite the angle C.

So 5 must go in the correct spot. But sides 6 and 7 are more flexible. The correct equation for this triangle can be either of these:

52=62+722(6)(7)cosCor52=72+622(7)(6)cosC

To fix the equation, swap 5 with 6.


Submit your answer as:

ID is: 4393 Seed is: 2019

Cos rule proof

Adapted from DBE Nov 2016, Grade 11, P2, Q7
Maths formulas

In the diagram below ΔPQR is drawn with obtuse angle R^.

Prove that r2=p2+q22pqcosR^.

  1. To prove r2=p2+q22pqcosR^, we need to make a construction. Which of the following represents a correct construction for this proof?

    I II
    III IV

    Using the construction, select the correct expression for the length of RS and QS.

    TIP: On a piece of paper, draw the construction diagram so that you can work from it. Name the sides of the triangle in the construction using the lower case letter of the opposite angle, eg. Side a opposite A^.
    Answer:

    The correct construction is Option:

    RS =

    QS =

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The correct construction will give us a right-angled triangle so that we can find an expression for the lengths using trig ratios.


    STEP: Choose a correct construction
    [−1 point ⇒ 1 / 2 points left]

    We need to construct a perpendicular height of the triangle PQR. Then we can write out two expressions using Pythagoras' theorem.

    A correct construction for this proof is:

    Option III is correct.


    STEP: Write an expression for cosR^ and sinR^
    [−1 point ⇒ 0 / 2 points left]

    Use the right-angled ΔQSR to write an expression for cos(180°R^) and sin(180°R^).

    cos(180°R^)=RSpcosR^=RSpRS=pcosR^
    sin(180°R^)=QSpQS=psinR^

    So, the correct expression for RS is pcosR^ and the correct expression for QS is psinR^.


    Submit your answer as: andand
  2. Nkechi was asked to prove r2=p2+q22pqcosR^.

    Using the correct construction, she labelled the length RS as x and wrote out the correct proof.

    TIP: On a piece of paper, draw this construction and write out the cos rule proof for yourself.

    Select the correct missing pieces so that Nkechi's proof is complete.

    Answer:
    Proof Steps
    1. =xp
    2. x=pcosR^
    In ΔQPS
    3. r2=QS2+(Pythagoras)
    4. QS2=r2q22qxx2
    In ΔQSR
    5. p2=QS2+x2(Pythagoras)
    6. QS2=
    7. r2q22qxx2=p2x2
    8. r2=q2+p2+
    9. r2=p2+q22pqcosR^
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    In a proof you need to show each step. The steps also need to be written out in the correct order. Read through Nkechi's steps and select the correct answers needed for this to be a correct proof.


    STEP: Prove the cos rule
    [−4 points ⇒ 0 / 4 points left]

    In a proof you need to show each step. The steps also need to be written out in the correct order.

    Start by labelling all the information you know on the diagram with the correct construction.

    We are going to prove that: r2=p2+q22pqcosR^.

    Here are the steps that are needed:

    1. Let RS=x and write an expression for cos(180°R^) in terms of x.
    2. Make x the subject of the expression.
    3. Use Pythagoras to write an expression for the sides in ΔQPS.
    4. Make QS2 the subject.
    5. Use Pythagoras to write an expression for the sides in ΔQSR.
    6. Make QS2 the subject.
    7. You have two expressions for QS2. Equate these.
    8. Make r2 the subject.
    9. Substitute in the expression for x to complete the cos rule.

    The completed proof is shown in the table below:

    Proof Steps
    1. cos(180°R^)=xp
    2. x=pcosR^
    In ΔQPS
    3. r2=QS2+(q+x)2(Pythagoras)
    4. QS2=r2q22qxx2
    In ΔQSR
    5. p2=QS2+x2(Pythagoras)
    6. QS2=p2x2
    7. r2q22qxx2=p2x2
    8. r2=q2+p2+2qx
    9. r2=p2+q22pqcosR^

    Submit your answer as: andandand

ID is: 4393 Seed is: 4577

Cos rule proof

Adapted from DBE Nov 2016, Grade 11, P2, Q7
Maths formulas

In the diagram below ΔABC is drawn with obtuse angle C^.

Prove that c2=a2+b22abcosC^.

  1. To prove c2=a2+b22abcosC^, we need to make a construction. Which of the following represents a correct construction for this proof?

    I II
    III IV

    Using the construction, select the correct expression for the y-coordinate of B and the x-coordinate of D.

    TIP: On a piece of paper, draw the construction diagram so that you can work from it. Name the sides of the triangle in the construction using the lower case letter of the opposite angle, eg. Side a opposite A^.
    Answer:

    The correct construction is Option:

    y-coordinate of B =

    x-coordinate of D =

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The correct construction will give us a right-angled triangle so that we can find an expression for the lengths using trig ratios.


    STEP: Choose a correct construction
    [−1 point ⇒ 1 / 2 points left]

    We need to construct a perpendicular line that will help us to find the coordinates of Point B.

    A correct construction for this proof is:

    Option I is correct.


    STEP: Write an expression for cosC^ and sinC^
    [−1 point ⇒ 0 / 2 points left]

    Let Point B be named (Bx,By) and Point D be named (Dx,Dy). Then we can use Points B and D to write an expression for sinC^ and cosC^.

    sinC^=ByaBy=asinC^
    cosC^=Bxa=DxaDx=acosC^

    So, the y-coordinate of B is asinC^ and the x-coordinate of D is acosC^.


    Submit your answer as: andand
  2. Sehlolo was asked to prove c2=a2+b22abcosC^.

    Using the correct construction, he labelled the coordinates of A, B and C, and wrote out the correct proof.

    TIP: On a piece of paper, draw this construction and write out the cos rule proof for yourself.

    Select the correct missing pieces so that Sehlolo's proof is complete.

    Answer:
    Proof Steps
    1. A(b;0)B(;asinC^)C(0;0)
    2. AB2=AD2+BD2(Pythagoras)
    3. c2=(bacosC^)2+
    4. c2=b22abcosC^+a2cos2C^+
    5. c2=b22abcosC^+a2(cos2C^+sin2C^)
    6. c2=b22abcosC^+a2()
    7. c2=a2+b22abcosC^
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    In a proof you need to show each step. The steps also need to be written out in the correct order. Read through Sehlolo's steps and select the correct answers needed for this to be a correct proof.


    STEP: Prove the cos rule
    [−4 points ⇒ 0 / 4 points left]

    In a proof you need to show each step. The steps also need to be written out in the correct order.

    Start by labelling all the information you know on the diagram with the correct construction.

    We are going to prove that: c2=a2+b22abcosC^.

    Here are the steps that are needed:

    1. Show the coordinates of Points A, B and C on your diagram and list these in the proof.
    2. Use Pythagoras to write an expression for the sides in ΔABD.
    3. Use the distance formula to find the distance AB2.
    4. Expand the expression by squaring out the brackets.
    5. Take out the common factor of a2.
    6. Give the simplified expression for the square identity.
    7. Reorder terms to match the cos rule statement.

    The completed proof is shown in the table below:

    Proof Steps
    1. A(b;0)B(acosC^;asinC^)C(0;0)
    2. AB2=AD2+BD2(Pythagoras)
    3. c2=(bacosC^)2+(0asinC^)2
    4. c2=b22abcosC^+a2cos2C^+a2sin2C^
    5. c2=b22abcosC^+a2(cos2C^+sin2C^)
    6. c2=b22abcosC^+a2(1)
    7. c2=a2+b22abcosC^

    Submit your answer as: andandand

ID is: 4393 Seed is: 5655

Cos rule proof

Adapted from DBE Nov 2016, Grade 11, P2, Q7
Maths formulas

In the diagram below ΔPQR is drawn with obtuse angle R^.

Prove that r2=p2+q22pqcosR^.

  1. To prove r2=p2+q22pqcosR^, we need to make a construction. Which of the following represents a correct construction for this proof?

    I II
    III IV

    Using the construction, select the correct expression for the y-coordinate of Q and the x-coordinate of S.

    TIP: On a piece of paper, draw the construction diagram so that you can work from it. Name the sides of the triangle in the construction using the lower case letter of the opposite angle, eg. Side a opposite A^.
    Answer:

    The correct construction is Option:

    y-coordinate of Q =

    x-coordinate of S =

    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The correct construction will give us a right-angled triangle so that we can find an expression for the lengths using trig ratios.


    STEP: Choose a correct construction
    [−1 point ⇒ 1 / 2 points left]

    We need to construct a perpendicular line that will help us to find the coordinates of Point Q.

    A correct construction for this proof is:

    Option IV is correct.


    STEP: Write an expression for cosR^ and sinR^
    [−1 point ⇒ 0 / 2 points left]

    Let Point Q be named (Qx,Qy) and Point S be named (Sx,Sy). Then we can use Points Q and S to write an expression for sinR^ and cosR^.

    sinR^=QypQy=psinR^
    cosR^=Qxp=SxpSx=pcosR^

    So, the y-coordinate of Q is psinR^ and the x-coordinate of S is pcosR^.


    Submit your answer as: andand
  2. Ntombikanina was asked to prove r2=p2+q22pqcosR^.

    Using the correct construction, she labelled the coordinates of P, Q and R, and wrote out the correct proof.

    TIP: On a piece of paper, draw this construction and write out the cos rule proof for yourself.

    Select the correct missing pieces so that Ntombikanina's proof is complete.

    Answer:
    Proof Steps
    1. P(q;0)Q(;psinR^)R(0;0)
    2. PQ2=PS2+QS2(Pythagoras)
    3. r2=(qpcosR^)2+
    4. r2=q22pqcosR^+p2cos2R^+
    5. r2=q22pqcosR^+p2(cos2R^+sin2R^)
    6. r2=q22pqcosR^+p2()
    7. r2=p2+q22pqcosR^
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    In a proof you need to show each step. The steps also need to be written out in the correct order. Read through Ntombikanina's steps and select the correct answers needed for this to be a correct proof.


    STEP: Prove the cos rule
    [−4 points ⇒ 0 / 4 points left]

    In a proof you need to show each step. The steps also need to be written out in the correct order.

    Start by labelling all the information you know on the diagram with the correct construction.

    We are going to prove that: r2=p2+q22pqcosR^.

    Here are the steps that are needed:

    1. Show the coordinates of Points P, Q and R on your diagram and list these in the proof.
    2. Use Pythagoras to write an expression for the sides in ΔPQS.
    3. Use the distance formula to find the distance PQ2.
    4. Expand the expression by squaring out the brackets.
    5. Take out the common factor of p2.
    6. Give the simplified expression for the square identity.
    7. Reorder terms to match the cos rule statement.

    The completed proof is shown in the table below:

    Proof Steps
    1. P(q;0)Q(pcosR^;psinR^)R(0;0)
    2. PQ2=PS2+QS2(Pythagoras)
    3. r2=(qpcosR^)2+(0psinR^)2
    4. r2=q22pqcosR^+p2cos2R^+p2sin2R^
    5. r2=q22pqcosR^+p2(cos2R^+sin2R^)
    6. r2=q22pqcosR^+p2(1)
    7. r2=p2+q22pqcosR^

    Submit your answer as: andandand

ID is: 3590 Seed is: 8501

Using the cosine rule: is there enough information?

The triangle below has a side with a length of 7. One of the angles is 46° and another angle is 75°. One of the angles is missing and it is labelled '?'.

Answer the two questions below about this diagram and the cosine rule.

Answer:
  1. Can you find the missing angle using the cosine rule?
  2. Why / why not?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Compare the information given about the triangle to the cosine rule. Is there enough information to use the cosine rule?


STEP: Compare the information given to the information used in the cosine rule
[−1 point ⇒ 1 / 2 points left]

This question is about the information in the cosine rule. The cosine rule connects all three sides of a triangle and one angle:

c2=a2+b22abcosθ

In this equation, a, b, and c are the sides of the triangle and θ is the angle opposite from side c in the triangle.

In total, the cosine rule contains four pieces of information about a triangle: all three sides and one angle. If we know any three of those four values we can use the cosine rule to calculate the fourth value. But if we have fewer than three of those values we do not have enough information to use the cosine rule.

In this case the cosine rule can't do the job: we cannot find the missing angle using the cosine rule unless we know the other two sides of the triangle. In other words, we only know one of the four values in the cosine rule. So it is not possible to calculate the value of the unknown angle with the cosine rule.

NOTE: It is possible to find the third angle because the three angles have a sum of 180°. But we cannot find the value using the cosine rule.

STEP: Select the correct answers
[−1 point ⇒ 0 / 2 points left]

Based on the reasons above we can select the correct answer choices:

  1. Can you find the missing angle using the cosine rule? No
  2. Why / why not? There is not enough information given.

Submit your answer as: and

ID is: 3590 Seed is: 3223

Using the cosine rule: is there enough information?

The triangle below has sides with lengths 6, 7, and 8. One of the angles is missing and it is labelled '?'.

Answer the two questions below about this diagram and the cosine rule.

Answer:
  1. Can you find the missing angle using the cosine rule?
  2. Why / why not?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Compare the information given about the triangle to the cosine rule. Is there enough information to use the cosine rule?


STEP: Compare the information given to the information used in the cosine rule
[−1 point ⇒ 1 / 2 points left]

This question is about the information in the cosine rule. The cosine rule connects all three sides of a triangle and one angle:

c2=a2+b22abcosθ

In this equation, a, b, and c are the sides of the triangle and θ is the angle opposite from side c in the triangle.

In total, the cosine rule contains four pieces of information about a triangle: all three sides and one angle. If we know any three of those four values we can use the cosine rule to calculate the fourth value. But if we have fewer than three of those values we do not have enough information to use the cosine rule.

In this case we know the values for a, b, and c in the cosine rule, and the only thing missing is the angle. So we can use the cosine rule to calculate the value of the unknown angle.


STEP: Select the correct answers
[−1 point ⇒ 0 / 2 points left]

Based on the reasons above we can select the correct answer choices:

  1. Can you find the missing angle using the cosine rule? Yes
  2. Why / why not? We know 3 of the 4 things in the cosine rule.

Submit your answer as: and

ID is: 3590 Seed is: 2970

Using the cosine rule: is there enough information?

The triangle below has two sides with lengths 7 and 6. The angle between those sides is 44°. One of the angles is missing and it is labelled '?'.

Answer the two questions below about this diagram and the cosine rule.

Answer:
  1. Can you find the missing angle using the cosine rule?
  2. Why / why not?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Compare the information given about the triangle to the cosine rule. Is there enough information to use the cosine rule?


STEP: Compare the information given to the information used in the cosine rule
[−1 point ⇒ 1 / 2 points left]

This question is about the information in the cosine rule. The cosine rule connects all three sides of a triangle and one angle:

c2=a2+b22abcosθ

In this equation, a, b, and c are the sides of the triangle and θ is the angle opposite from side c in the triangle.

In total, the cosine rule contains four pieces of information about a triangle: all three sides and one angle. If we know any three of those four values we can use the cosine rule to calculate the fourth value. But if we have fewer than three of those values we do not have enough information to use the cosine rule.

In this case the cosine rule can't do the job: we can only find an angle using the cosine rule if we know all three sides of the triangle. In other words, we only know two of the four values in the cosine rule. So it is not possible to calculate the value of the unknown angle with the cosine rule.

NOTE: It is possible to find the missing angle using the sine rule, which you might already have learned. But we cannot find that angle using the cosine rule.

STEP: Select the correct answers
[−1 point ⇒ 0 / 2 points left]

Based on the reasons above we can select the correct answer choices:

  1. Can you find the missing angle using the cosine rule? No
  2. Why / why not? There is not enough information given.

Submit your answer as: and

ID is: 3503 Seed is: 9645

Finding an angle with the cosine rule

The triangle below has all three sides labelled. CB¯=8 units, BA¯=7 units, and CA¯=6 units. The figure is drawn to scale.

Find the measure of the angle C^.

TIP: The question states that the figure is drawn to scale. And the angle that you want is acute. In fact, it must be quite a bit less than 90°. So make sure your answer agrees with that!
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer:

The measure of angle C^ is °.

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by writing the cosine rule equation. Then substitute in the lengths for the sides of the triangle.


STEP: Identify the side and angle opposite pair for the cosine rule
[−1 point ⇒ 4 / 5 points left]

We need to find the size of angle C^ in the triangle. We can find this angle using the cosine rule. The cosine rule relates all three sides of the triangle and one of the angles.

c2=a2+b22abcosθ

To use this equation, we must be sure that the side we use for c in the formula is the side of the triangle across from (opposite) angle θ. These opposites are shown below:

So in this case c=7 because that is the side of the triangle opposite from angle C^.


STEP: Substitute into the formula
[−1 point ⇒ 3 / 5 points left]

We can organise the information that we need for the cosine rule:

  • We want to find the angle, so we will use x in place of θ (C^ would also be a good choice).
  • c is the side across from the angle, so c=7.
  • a and b are the other sides of the triangle. It does not matter which side we use for a and which for b.

Substitute all these values into the equation:

c2=a2+b22abcosθ(7)2=(6)2+(8)22(6)(8)cosx

STEP: Solve for the angle
[−2 points ⇒ 1 / 5 points left]

Now we need to do all of the calculations required to solve for x.

(7)2=(6)2+(8)22(6)(8)cosx49=36+64(96)cosx49=10096cosx51=96cosx5196=cosx1732=cosx

At this point we must use the inverse cosine function, cos1. When we use cos1, we will get one answer from the calculator. But we must remember that there is a second solution, which is the negative of what we get from the calculator (because cosx=cos(x)). And in addition to that, we must add +360°k to each solution.

1732=cosxcos1(1732)=cos1(cosx)This produces two solutions:x=57.91004...°+360°k,kZorx=57.91004...°+360°k,kZ

STEP: Determine the final answer
[−1 point ⇒ 0 / 5 points left]

To finish the solution we need to use the fact that we are solving for an angle in a triangle. That means two things: (a) the answer cannot be negative because angles in a triangle cannot be negative numbers, and (b) the answer cannot be larger than 180° because the angles in a triangle add up to 180° (they cannot be larger than that).

That means we can ignore the negative angle (in the second solution above). And we can also ignore the 360°k terms. That leaves just the one acceptable answer:

x=57.91004...°

The instructions told us to round to two decimal places. So: 57.91004...°57.91°. Remember to compare your answer to the diagram: is 57.91° a reasonable answer for C^?

Therefore the correct answer is C^=57.91°.


Submit your answer as:

ID is: 3503 Seed is: 9256

Finding an angle with the cosine rule

The triangle below has all three sides labelled. BA¯=9 units, AC¯=6 units, and BC¯=5 units. The figure is drawn to scale.

Determine the measure of the angle A^.

TIP: The question states that the figure is drawn to scale. And the angle that you want is acute. In fact, it must be quite a bit less than 90°. So make sure your answer agrees with that!
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer:

The measure of angle A^ is °.

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by writing the cosine rule equation. Then substitute in the lengths for the sides of the triangle.


STEP: Identify the side and angle opposite pair for the cosine rule
[−1 point ⇒ 4 / 5 points left]

We need to find the size of angle A^ in the triangle. We can find this angle using the cosine rule. The cosine rule relates all three sides of the triangle and one of the angles.

c2=a2+b22abcosθ

To use this equation, we must be sure that the side we use for c in the formula is the side of the triangle across from (opposite) angle θ. These opposites are shown below:

So in this case c=5 because that is the side of the triangle opposite from angle A^.


STEP: Substitute into the formula
[−1 point ⇒ 3 / 5 points left]

We can organise the information that we need for the cosine rule:

  • We want to find the angle, so we will use x in place of θ (A^ would also be a good choice).
  • c is the side across from the angle, so c=5.
  • a and b are the other sides of the triangle. It does not matter which side we use for a and which for b.

Substitute all these values into the equation:

c2=a2+b22abcosθ(5)2=(9)2+(6)22(9)(6)cosx

STEP: Solve for the angle
[−2 points ⇒ 1 / 5 points left]

Now we need to do all of the calculations required to solve for x.

(5)2=(9)2+(6)22(9)(6)cosx25=81+36(108)cosx25=117108cosx92=108cosx92108=cosx2327=cosx

At this point we must use the inverse cosine function, cos1. When we use cos1, we will get one answer from the calculator. But we must remember that there is a second solution, which is the negative of what we get from the calculator (because cosx=cos(x)). And in addition to that, we must add +360°k to each solution.

2327=cosxcos1(2327)=cos1(cosx)This produces two solutions:x=31.58633...°+360°k,kZorx=31.58633...°+360°k,kZ

STEP: Determine the final answer
[−1 point ⇒ 0 / 5 points left]

To finish the solution we need to use the fact that we are solving for an angle in a triangle. That means two things: (a) the answer cannot be negative because angles in a triangle cannot be negative numbers, and (b) the answer cannot be larger than 180° because the angles in a triangle add up to 180° (they cannot be larger than that).

That means we can ignore the negative angle (in the second solution above). And we can also ignore the 360°k terms. That leaves just the one acceptable answer:

x=31.58633...°

The instructions told us to round to two decimal places. So: 31.58633...°31.59°. Remember to compare your answer to the diagram: is 31.59° a reasonable answer for A^?

Therefore the correct answer is A^=31.59°.


Submit your answer as:

ID is: 3503 Seed is: 7334

Finding an angle with the cosine rule

The triangle below has all three sides labelled. AC¯=7 units, CB¯=5 units, and AB¯=9 units. The figure is drawn to scale.

Determine the measure of the angle C^.

TIP: The question states that the figure is drawn to scale. And the angle that you want is obtuse (greater than 90°). So make sure your answer agrees with that!
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer:

The measure of angle C^ is °.

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by writing the cosine rule equation. Then substitute in the lengths for the sides of the triangle.


STEP: Identify the side and angle opposite pair for the cosine rule
[−1 point ⇒ 4 / 5 points left]

We need to find the size of angle C^ in the triangle. We can find this angle using the cosine rule. The cosine rule relates all three sides of the triangle and one of the angles.

c2=a2+b22abcosθ

To use this equation, we must be sure that the side we use for c in the formula is the side of the triangle across from (opposite) angle θ. These opposites are shown below:

So in this case c=9 because that is the side of the triangle opposite from angle C^.


STEP: Substitute into the formula
[−1 point ⇒ 3 / 5 points left]

We can organise the information that we need for the cosine rule:

  • We want to find the angle, so we will use x in place of θ (C^ would also be a good choice).
  • c is the side across from the angle, so c=9.
  • a and b are the other sides of the triangle. It does not matter which side we use for a and which for b.

Substitute all these values into the equation:

c2=a2+b22abcosθ(9)2=(7)2+(5)22(7)(5)cosx

STEP: Solve for the angle
[−2 points ⇒ 1 / 5 points left]

Now we need to do all of the calculations required to solve for x.

(9)2=(7)2+(5)22(7)(5)cosx81=49+25(70)cosx81=7470cosx7=70cosx770=cosx110=cosx

At this point we must use the inverse cosine function, cos1. When we use cos1, we will get one answer from the calculator. But we must remember that there is a second solution, which is the negative of what we get from the calculator (because cosx=cos(x)). And in addition to that, we must add +360°k to each solution.

110=cosxcos1(110)=cos1(cosx)This produces two solutions:x=95.73917...°+360°k,kZorx=95.73917...°+360°k,kZ

STEP: Determine the final answer
[−1 point ⇒ 0 / 5 points left]

To finish the solution we need to use the fact that we are solving for an angle in a triangle. That means two things: (a) the answer cannot be negative because angles in a triangle cannot be negative numbers, and (b) the answer cannot be larger than 180° because the angles in a triangle add up to 180° (they cannot be larger than that).

That means we can ignore the negative angle (in the second solution above). And we can also ignore the 360°k terms. That leaves just the one acceptable answer:

x=95.73917...°

The instructions told us to round to two decimal places. So: 95.73917...°95.74°. Remember to compare your answer to the diagram: is 95.74° a reasonable answer for C^?

Therefore the correct answer is C^=95.74°.


Submit your answer as:

ID is: 1555 Seed is: 2880

Using the cosine rule

Calculate the length of line segment ZX¯ in the triangle below. The triangle has sides ZY¯ and YX¯ equal to 10.4 and 8 units respectively. Also Y^=64°. The figure is drawn to scale.

TIP: The question states that the figure is drawn to scale. So if the unknown side looks longer than the other sides in the triangle, your answer must be larger that the sides shown. If it looks shorter, your answer must be smaller. And so on.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer: Side ZX¯ is units.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by writing down the cosine rule formula. Make sure that you note the side opposite the angle so that you can substitute into the formula correctly.


STEP: Identify the side which is opposite the angle (for the cosine rule)
[−1 point ⇒ 3 / 4 points left]

We can answer this question using the cosine rule, which says:

c2=a2+b22abcosθ

where c is the side of the triangle opposite to angle θ. We can identify these opposites on the diagram like this:

In this case, the angle is 64° and the missing side is the value of c (in the cosine rule).


STEP: Substitute into the formula and do the calculations
[−2 points ⇒ 1 / 4 points left]

Now we can substitute the values into the cosine rule and solve. Note these useful facts:

  • For the two non-opposite sides, it does not matter which is a and which is b.
  • We can use x in place of the “?”, because that is more familiar.
c2=a2+b22abcosθx2=(10.4)2+(8)22(10.4)(8)cos(64°)x2=108.16+64(166.4)(0.43837...)x2=172.1672.94495...x2=99.21504...

STEP: Complete the solution to find the side
[−1 point ⇒ 0 / 4 points left]

Now square root both sides to remove the square on x.

x2=99.21504...of an equation brings in a ±.Square rooting both sidesx=±99.21504...x=±9.96067...

There are two numerical answers for this question: 9.96067... and 9.96067.... But the length cannot be negative, so we must throw out the negative solution. Remember to check the answer against the diagram (we can do that because the diagram is drawn to scale). In this case, our answer is greater than one of the other sides of the triangle and less than the other side. Is the side we wanted longer than one of the other sides but shorter than the third side?

Therefore, the correct answer, rounded to two places, is: x=9.96 units.


Submit your answer as:

ID is: 1555 Seed is: 7086

Using the cosine rule

Calculate the length of line segment BA¯ in the triangle below. The triangle has sides BC¯ and CA¯ equal to 10.7 and 13.1 units respectively. Also C^=69°. The figure is drawn to scale.

TIP: The question states that the figure is drawn to scale. So if the unknown side looks longer than the other sides in the triangle, your answer must be larger that the sides shown. If it looks shorter, your answer must be smaller. And so on.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer: Side BA¯ is units.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by writing down the cosine rule formula. Make sure that you note the side opposite the angle so that you can substitute into the formula correctly.


STEP: Identify the side which is opposite the angle (for the cosine rule)
[−1 point ⇒ 3 / 4 points left]

We can answer this question using the cosine rule, which says:

c2=a2+b22abcosθ

where c is the side of the triangle opposite to angle θ. We can identify these opposites on the diagram like this:

In this case, the angle is 69° and the missing side is the value of c (in the cosine rule).


STEP: Substitute into the formula and do the calculations
[−2 points ⇒ 1 / 4 points left]

Now we can substitute the values into the cosine rule and solve. Note these useful facts:

  • For the two non-opposite sides, it does not matter which is a and which is b.
  • We can use x in place of the “?”, because that is more familiar.
c2=a2+b22abcosθx2=(10.7)2+(13.1)22(10.7)(13.1)cos(69°)x2=114.49+171.61(280.34)(0.35836...)x2=286.1100.46487...x2=185.63512...

STEP: Complete the solution to find the side
[−1 point ⇒ 0 / 4 points left]

Now square root both sides to remove the square on x.

x2=185.63512...of an equation brings in a ±.Square rooting both sidesx=±185.63512...x=±13.62479...

There are two numerical answers for this question: 13.62479... and 13.62479.... But the length cannot be negative, so we must throw out the negative solution. Remember to check the answer against the diagram (we can do that because the diagram is drawn to scale). In this case, our answer is larger than the other sides of the triangle. Is the side we wanted the longest side of the triangle?

Therefore, the correct answer, rounded to two places, is: x=13.62 units.


Submit your answer as:

ID is: 1555 Seed is: 9145

Using the cosine rule

Determine the value of Z^ in the triangle shown. The sides of the triangle have lengths as follows: YZ¯=7.1 units, ZX¯=13.4 units, and YX¯=11.9 units. The figure is drawn to scale.

TIP: The question states that the figure is drawn to scale. And the angle that you want is acute (less than 90°). So make sure your answer agrees with that!
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer: Angle Z^ is °.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by writing down the cosine rule formula. Make sure that you note the side opposite the angle so that you can substitute into the formula correctly.


STEP: Identify the side which is opposite the angle (for the cosine rule)
[−1 point ⇒ 3 / 4 points left]

We can answer this question using the cosine rule, which says:

c2=a2+b22abcosθ

where c is the side of the triangle opposite to angle θ. We can identify these opposites on the diagram like this:

In this case, the side 11.9 units long is opposite the angle we need to find. So we will use c=11.9 in the cosine rule.


STEP: Substitute into the formula and do the calculations
[−2 points ⇒ 1 / 4 points left]

Now we can substitute the values into the cosine rule and solve. Note these useful facts:

  • For the two non-opposite sides, it does not matter which is a and which is b.
  • We can use x in place of the “?°” (Z^ would also be a good choice).
c2=a2+b22abcosθ(11.9)2=(7.1)2+(13.4)22(7.1)(13.4)cosx141.61=50.41+179.56(190.28)cosx141.61=229.97190.28cosx88.36=190.28cosx88.36190.28=cosx0.464368...=cosx

STEP: Complete the solution to find the angle
[−1 point ⇒ 0 / 4 points left]

Now we need to solve for an angle. So we must use the inverse cosine function. Remember that when we use cos1(something) we must bring in the +360°k and the second solution! However, we can ignore the +360°k term in this case because the angle we want is inside of a triangle: it cannot be greater than 180°. That leaves the second solution, which is the opposite of whatever answer you get from your calculator.

0.464368...=cosxcos1(0.464368...)=cos1(cosx)62.33065...°=xor62.33065...°=x

There are now two answers. But an angle in a triangle cannot be negative! So we should ignore the negative solution. Remember to check the answer against the diagram (we can do that because the diagram is drawn to scale). Does the missing angle in the diagram look like it is about 62°?

The instructions told us to round to two decimal places. Therefore the measure of angle Z^ is: x=62.33°.


Submit your answer as:

ID is: 1432 Seed is: 5252

The cosine rule: identifying the correct sides and angles

The triangle below has sides labelled m,n, and p, and angles labelled θ,α, and β.

The cosine rule states:

p2=n2+m22nmcos(?)

Which of the following choices belongs in the place of the '?' in the formula?

Answer: The missing quantity is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

When using the cosine rule, you must remember that the angle used in the cos function must be opposite from the side on the other side of the equal sign in the formula. For example, if one writes h2=j2+k22jkcosG^, then side h must be across the triangle from the angle G^.


STEP: Connect the sides and angles in the diagram according to the cosine rule
[−1 point ⇒ 0 / 1 points left]

The cosine rule has three sides and one angle in it. We must make sure that the sides and angles go in the correct places in the formula! Specifically, the side which is opposite the angle in the triangle must go on the opposite side of the equation from the angle.

In this question, we are looking for one of the angles:

p2=n2+m22nmcos(?)

This angle must be the angle across from side p. Looking at the figure, we can see that this must be angle β.

The correct choice from the list is β.


Submit your answer as:

ID is: 1432 Seed is: 998

The cosine rule: identifying the correct sides and angles

The triangle below has sides labelled a,b, and c, and angles labelled A,B, and C.

The cosine rule states:

b2=c2+a22cacos(?)

Which of the following choices belongs in the place of the '?' in the formula?

Answer: The missing quantity is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

When using the cosine rule, you must remember that the angle used in the cos function must be opposite from the side on the other side of the equal sign in the formula. For example, if one writes h2=j2+k22jkcosG^, then side h must be across the triangle from the angle G^.


STEP: Connect the sides and angles in the diagram according to the cosine rule
[−1 point ⇒ 0 / 1 points left]

The cosine rule has three sides and one angle in it. We must make sure that the sides and angles go in the correct places in the formula! Specifically, the side which is opposite the angle in the triangle must go on the opposite side of the equation from the angle.

In this question, we are looking for one of the angles:

b2=c2+a22cacos(?)

This angle must be the angle across from side b. Looking at the figure, we can see that this must be angle B.

The correct choice from the list is B.


Submit your answer as:

ID is: 1432 Seed is: 7433

The cosine rule: identifying the correct sides and angles

The triangle below has sides labelled c,a, and b, and angles labelled C,A, and B.

The cosine rule states:

c2=(?)2+b22(?)bcosC

Which of the following choices belongs in the place of the '?' in the formula?

Answer: The missing quantity is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

When using the cosine rule, you must remember that the angle used in the cos function must be opposite from the side on the other side of the equal sign in the formula. For example, if one writes h2=j2+k22jkcosG^, then side h must be across the triangle from the angle G^.


STEP: Connect the sides and angles in the diagram according to the cosine rule
[−1 point ⇒ 0 / 1 points left]

The cosine rule has three sides and one angle in it. We must make sure that the sides and angles go in the correct places in the formula! Specifically, the side which is opposite the angle in the triangle must go on the opposite side of the equation from the angle.

In this question, we are looking for one of the sides:

c2=(?)2+b22(?)bcosC

The formula already includes both the angle C and the side across from it, c. That leaves the other two sides of the triangle, a and b. Since b is already in the formula, the answer must be a.

The correct choice from the list is a.


Submit your answer as:

ID is: 3591 Seed is: 3393

Cosine rule connections

  1. The cosine rule is about triangles, including right-angled triangles. And if you use a 90° angle in the cosine rule, it becomes the same as another important equation. Which statement below identifies what happens if we use a 90° angle in the cosine rule?

    Choices If you use a 90° angle is the cosine rule...
    A the sum of the interior angles of a triangle is 180°
    B it becomes the theorem of Pythagoras
    C it changes into the sine rule
    D it changes into the area rule
    Answer: The correct statement is choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Draw a triangle to represent the situation described in the question: a right-angled triangle. Then work out the cosine rule including the right-angle.


    STEP: Connect the cosine rule to the theorem of Pythagoras
    [−1 point ⇒ 0 / 1 points left]

    The cosine rule works for any triangle. But suppose that there is a 90° angle in the triangle. Then we have a right-angled triangle.

    Since this is a right-angled triangle, the sides obey the theorem of Pythagoras. But the sides also obey the cosine rule. And if we include the right-angle when we write the cosine rule, it ends up being equivalent to the theorem of Pythagoras. (Question 2 is about why this happens.)

    Using a 90° angle in the cosine rule gives us the theorem of Pythagoras.


    Submit your answer as:
  2. The table below contains a proof. The proof shows that the cosine rule simplifies to the theorem of Pythagoras if the angle in the cosine rule is 90°. However, there is one step and one reason missing. Select the correct option from each list of choices to complete the proof.

    Answer:
    Steps Reasons
    Given
    c2=a2+b22abcosC^
    Substitute C^=90°
    c2=a2+b22ab(0) Evaluate cos(90°)=0
    c2=a2+b2 Multiply by 0
    Theorem of Pythagoras applies to ΔABC c is the hypotenuse, a and b are legs
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by analysing the steps and reasons which are already shown. Think about what they mean. Then try to decide what the missing pieces should be.


    STEP: Select the correct step and reason
    [−2 points ⇒ 0 / 2 points left]

    The proof in the question connects the cosine rule to right-angled triangles. If we substitute 90° for θ, the cosine rule becomes the same as the theorem of Pythagoras. In other words, the cosine rule ends up becoming equivalent to the equation c2=a2+b2. This connection is based on the fact that cos90° is equal to 0.

    The table below shows the complete proof. The correct answers are in the shaded blocks.

    Steps Reasons
    Given
    c2=a2+b22abcosC^ Cosine rule using C^
    c2=a2+b22abcos(90°) Substitute C^=90°
    c2=a2+b22ab(0) Evaluate cos(90°)=0
    c2=a2+b2 Multiply by 0
    Theorem of Pythagoras applies to ΔABC c is the hypotenuse, a and b are legs

    Submit your answer as: and

ID is: 3591 Seed is: 9340

Cosine rule connections

  1. The cosine rule is about triangles, including right-angled triangles. And if you use a 90° angle in the cosine rule, it becomes the same as another important equation. Which statement below identifies what happens if we use a 90° angle in the cosine rule?

    Choices If you use a 90° angle is the cosine rule...
    A it becomes the equilateral triangle calculation
    B it becomes the theorem of Pythagoras
    C it changes into the sine rule
    D the sum of the interior angles of a triangle is 180°
    Answer: The correct statement is choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Draw a triangle to represent the situation described in the question: a right-angled triangle. Then work out the cosine rule including the right-angle.


    STEP: Connect the cosine rule to the theorem of Pythagoras
    [−1 point ⇒ 0 / 1 points left]

    The cosine rule works for any triangle. But suppose that there is a 90° angle in the triangle. Then we have a right-angled triangle.

    Since this is a right-angled triangle, the sides obey the theorem of Pythagoras. But the sides also obey the cosine rule. And if we include the right-angle when we write the cosine rule, it ends up being equivalent to the theorem of Pythagoras. (Question 2 is about why this happens.)

    Using a 90° angle in the cosine rule gives us the theorem of Pythagoras.


    Submit your answer as:
  2. The table below contains a proof. The proof shows that the cosine rule simplifies to the theorem of Pythagoras if the angle in the cosine rule is 90°. However, there is one step and one reason missing. Select the correct option from each list of choices to complete the proof.

    Answer:
    Steps Reasons
    Given
    c2=a2+b22abcosC^ Cosine rule using C^
    c2=a2+b22abcos(90°) Substitute C^=90°
    Evaluate cos(90°)=0
    c2=a2+b2
    Theorem of Pythagoras applies to ΔABC c is the hypotenuse, a and b are legs
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by analysing the steps and reasons which are already shown. Think about what they mean. Then try to decide what the missing pieces should be.


    STEP: Select the correct step and reason
    [−2 points ⇒ 0 / 2 points left]

    The proof in the question connects the cosine rule to right-angled triangles. If we substitute 90° for θ, the cosine rule becomes the same as the theorem of Pythagoras. In other words, the cosine rule ends up becoming equivalent to the equation c2=a2+b2. This connection is based on the fact that cos90° is equal to 0.

    The table below shows the complete proof. The correct answers are in the shaded blocks.

    Steps Reasons
    Given
    c2=a2+b22abcosC^ Cosine rule using C^
    c2=a2+b22abcos(90°) Substitute C^=90°
    c2=a2+b22ab(0) Evaluate cos(90°)=0
    c2=a2+b2 Multiply by 0
    Theorem of Pythagoras applies to ΔABC c is the hypotenuse, a and b are legs

    Submit your answer as: and

ID is: 3591 Seed is: 3032

Cosine rule connections

  1. The cosine rule is about triangles, including right-angled triangles. And if you use a 90° angle in the cosine rule, it becomes the same as another important equation. Which statement below identifies what happens if we use a 90° angle in the cosine rule?

    Choices If you use a 90° angle is the cosine rule...
    A it becomes the isosceles triangle calculation
    B it changes into the sine rule
    C it becomes the theorem of Pythagoras
    D it changes into the area rule
    Answer: The correct statement is choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Draw a triangle to represent the situation described in the question: a right-angled triangle. Then work out the cosine rule including the right-angle.


    STEP: Connect the cosine rule to the theorem of Pythagoras
    [−1 point ⇒ 0 / 1 points left]

    The cosine rule works for any triangle. But suppose that there is a 90° angle in the triangle. Then we have a right-angled triangle.

    Since this is a right-angled triangle, the sides obey the theorem of Pythagoras. But the sides also obey the cosine rule. And if we include the right-angle when we write the cosine rule, it ends up being equivalent to the theorem of Pythagoras. (Question 2 is about why this happens.)

    Using a 90° angle in the cosine rule gives us the theorem of Pythagoras.


    Submit your answer as:
  2. The table below contains a proof. The proof shows that the cosine rule simplifies to the theorem of Pythagoras if the angle in the cosine rule is 90°. However, there is one step and one reason missing. Select the correct option from each list of choices to complete the proof.

    Answer:
    Steps Reasons
    Given
    c2=a2+b22abcosC^ Cosine rule using C^
    c2=a2+b22abcos(90°) Substitute C^=90°
    Evaluate cos(90°)=0
    c2=a2+b2
    Theorem of Pythagoras applies to ΔABC c is the hypotenuse, a and b are legs
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by analysing the steps and reasons which are already shown. Think about what they mean. Then try to decide what the missing pieces should be.


    STEP: Select the correct step and reason
    [−2 points ⇒ 0 / 2 points left]

    The proof in the question connects the cosine rule to right-angled triangles. If we substitute 90° for θ, the cosine rule becomes the same as the theorem of Pythagoras. In other words, the cosine rule ends up becoming equivalent to the equation c2=a2+b2. This connection is based on the fact that cos90° is equal to 0.

    The table below shows the complete proof. The correct answers are in the shaded blocks.

    Steps Reasons
    Given
    c2=a2+b22abcosC^ Cosine rule using C^
    c2=a2+b22abcos(90°) Substitute C^=90°
    c2=a2+b22ab(0) Evaluate cos(90°)=0
    c2=a2+b2 Multiply by 0
    Theorem of Pythagoras applies to ΔABC c is the hypotenuse, a and b are legs

    Submit your answer as: and

ID is: 3611 Seed is: 9144

Finding an adjacent side

The diagram below is roughly to scale. It shows triangle PQR. Side PQ is 13 units long and side PR is 11 units long. R^=θ. The value of cosθ=14. Determine the length of the third side, QR.

TIP: The question states that the figure is drawn roughly to scale. Make sure your answer is reasonable compared to the appearance of the figure.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer: Side QR¯ is units.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by writing down an equation for the triangle. Use the cosine rule.


STEP: Identify the side and angle opposite pair for the cosine rule
[−1 point ⇒ 5 / 6 points left]

The cosine rule says: c2=a2+b22abcosθ where c is the side of the triangle opposite to angle θ. So the first thing to do is identify the side of the triangle which is opposite from the angle of interest. In this question, the side we need is adjacent to the angle labelled as θ. It can be helpful to make these opposites more obvious, like this:


STEP: Substitute known values into the formula
[−2 points ⇒ 3 / 6 points left]

Now we can write a cosine rule equation. For the two sides which are not opposite to the angle, it does not matter which is a and which is b. Let's put x in place of the '?.'

c2=a2+b22abcosθ(13)2=(x)2+(11)22(x)(11)cosθ

Now substitute in the value of cosθ, which is equal to 14. We are subsituting here for the entire cosine factor, not the angle itself.

(13)2=(x)2+(11)22(x)(11)cosθ(13)2=(x)2+(11)22(x)(11)(14)

STEP: Arrange the equation into standard form
[−1 point ⇒ 2 / 6 points left]

This is a quadratic equation. We need to arrange it in standard form.

(13)2=(x)2+(11)22(x)(11)(14)169=x2+121112x0=x2112x+1211690=x2112x48

We can help ourselves by removing the fraction. To do that, multiply the equation (on both sides) by 2. It is important to see that this does not change the zero on the left side of the equation.

(2)0=(x2112x48)(2)0=2x211x96

STEP: Solve the equation with the quadratic formula
[−2 points ⇒ 0 / 6 points left]

Now use the quadratic formula to find the solutions (we can expet two solutions because the equation is quadratic).

x=b±b24ac2a=(11)±(11)24(2)(96)2(2)=11±121+7684=11±8894=11±29.81610...4

Now we can split the two answers apart (separate the plus-minus).

x=11+29.81610...4 or x=1129.81610...4=40.81610...4 or =18.81610...4=10.20402... or =4.70402...

We get two answers. But these values are for the side of a triangle, which is a distance. So we must throw out any negative answers. One of the answers above is negative, so we will drop it! That leaves one acceptable (positive) answer.

The correct answer, rounded to two places, is: x=10.2.


Submit your answer as:

ID is: 3611 Seed is: 1223

Finding an adjacent side

The diagram below is roughly to scale. It shows triangle ZYX. Side ZY is 19 units long and side ZX is 11 units long. X^=θ. The value of cosθ=14. Determine the length of the third side, YX.

TIP: The question states that the figure is drawn roughly to scale. Make sure your answer is reasonable compared to the appearance of the figure.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer: Side YX¯ is units.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by writing down an equation for the triangle. Use the cosine rule.


STEP: Identify the side and angle opposite pair for the cosine rule
[−1 point ⇒ 5 / 6 points left]

The cosine rule says: c2=a2+b22abcosθ where c is the side of the triangle opposite to angle θ. So the first thing to do is identify the side of the triangle which is opposite from the angle of interest. In this question, the side we need is adjacent to the angle labelled as θ. It can be helpful to make these opposites more obvious, like this:


STEP: Substitute known values into the formula
[−2 points ⇒ 3 / 6 points left]

Now we can write a cosine rule equation. For the two sides which are not opposite to the angle, it does not matter which is a and which is b. Let's put x in place of the '?.'

c2=a2+b22abcosθ(19)2=(x)2+(11)22(x)(11)cosθ

Now substitute in the value of cosθ, which is equal to 14. We are subsituting here for the entire cosine factor, not the angle itself.

(19)2=(x)2+(11)22(x)(11)cosθ(19)2=(x)2+(11)22(x)(11)(14)

STEP: Arrange the equation into standard form
[−1 point ⇒ 2 / 6 points left]

This is a quadratic equation. We need to arrange it in standard form.

(19)2=(x)2+(11)22(x)(11)(14)361=x2+121+112x0=x2+112x+1213610=x2+112x240

We can help ourselves by removing the fraction. To do that, multiply the equation (on both sides) by 2. It is important to see that this does not change the zero on the left side of the equation.

(2)0=(x2+112x240)(2)0=2x2+11x480

STEP: Solve the equation with the quadratic formula
[−2 points ⇒ 0 / 6 points left]

Now use the quadratic formula to find the solutions (we can expet two solutions because the equation is quadratic).

x=b±b24ac2a=(11)±(11)24(2)(480)2(2)=11±121+38404=11±39614=11±62.93647...4

Now we can split the two answers apart (separate the plus-minus).

x=11+62.93647...4 or x=1162.93647...4=51.93647...4 or =73.93647...4=12.98411... or =18.48411...

We get two answers. But these values are for the side of a triangle, which is a distance. So we must throw out any negative answers. One of the answers above is negative, so we will drop it! That leaves one acceptable (positive) answer.

The correct answer, rounded to two places, is: x=12.98.


Submit your answer as:

ID is: 3611 Seed is: 9702

Finding an adjacent side

The diagram below is roughly to scale. It shows triangle BAC. Side BA is 19 units long and side BC is 13 units long. C^=θ. The value of cosθ=112. Determine the length of the third side, AC.

TIP: The question states that the figure is drawn roughly to scale. Make sure your answer is reasonable compared to the appearance of the figure.
INSTRUCTION: Round your answer to two decimal places if necessary.
Answer: Side AC¯ is units.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by writing down an equation for the triangle. Use the cosine rule.


STEP: Identify the side and angle opposite pair for the cosine rule
[−1 point ⇒ 5 / 6 points left]

The cosine rule says: c2=a2+b22abcosθ where c is the side of the triangle opposite to angle θ. So the first thing to do is identify the side of the triangle which is opposite from the angle of interest. In this question, the side we need is adjacent to the angle labelled as θ. It can be helpful to make these opposites more obvious, like this:


STEP: Substitute known values into the formula
[−2 points ⇒ 3 / 6 points left]

Now we can write a cosine rule equation. For the two sides which are not opposite to the angle, it does not matter which is a and which is b. Let's put x in place of the '?.'

c2=a2+b22abcosθ(19)2=(x)2+(13)22(x)(13)cosθ

Now substitute in the value of cosθ, which is equal to 112. We are subsituting here for the entire cosine factor, not the angle itself.

(19)2=(x)2+(13)22(x)(13)cosθ(19)2=(x)2+(13)22(x)(13)(112)

STEP: Arrange the equation into standard form
[−1 point ⇒ 2 / 6 points left]

This is a quadratic equation. We need to arrange it in standard form.

(19)2=(x)2+(13)22(x)(13)(112)361=x2+169136x0=x2136x+1693610=x2136x192

We can help ourselves by removing the fraction. To do that, multiply the equation (on both sides) by 6. It is important to see that this does not change the zero on the left side of the equation.

(6)0=(x2136x192)(6)0=6x213x1152

STEP: Solve the equation with the quadratic formula
[−2 points ⇒ 0 / 6 points left]

Now use the quadratic formula to find the solutions (we can expet two solutions because the equation is quadratic).

x=b±b24ac2a=(13)±(13)24(6)(1152)2(6)=13±169+2764812=13±2781712=13±166.78429...12

Now we can split the two answers apart (separate the plus-minus).

x=13+166.78429...12 or x=13166.78429...12=179.78429...12 or =153.78429...12=14.98202... or =12.81535...

We get two answers. But these values are for the side of a triangle, which is a distance. So we must throw out any negative answers. One of the answers above is negative, so we will drop it! That leaves one acceptable (positive) answer.

The correct answer, rounded to two places, is: x=14.98.


Submit your answer as:

3. Practical applications


ID is: 3568 Seed is: 4816

Predator and prey

In the corner of a room, a chameleon and two flies are on the wall. The chameleon is hungry and hoping for a meal. The flies are 61.76 cm away from each other. The chameleon is 54.3 cm away from one of the flies and 57.52 cm away from the other fly.

  1. What is the angle subtended by the two flies at the chameleon's position?

    INSTRUCTION: Round to three decimal places.
    Answer: The angle is °.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by drawing a diagram. The shape you should use is a triangle. Then use the cosine rule to calculate the size of the angle you need.


    STEP: Draw a picture of the triangle
    [−1 point ⇒ 4 / 5 points left]

    We need to find an angle, but it is not obvious exactly which angle. We need to change the information in the question into a picture, so the first thing we need to do is draw a sketch. We should label it with everything we can.

    NOTE: The figure above is not drawn to scale. That is fine, but it means that we cannot draw any conclusions based on the appearance of the triangle. Only the labels are reliable.

    STEP: Identify the angle we want and write an equation using the cosine rule
    [−2 points ⇒ 2 / 5 points left]

    The question asks us to find the angle subtended by the two flies at the chameleon's position. The word subtends means 'make an angle at a point from two other points.' The angle subtended at the chameleon's position is the angle labelled chameleon. This is because that angle is formed by the lines going to the two other animals.

    Let's use a letter C for chameleon to represent the angle. Angle C is opposite from side 61.76 cm, so using the cosine rule we get:

    (61.76)2=(54.3)2+(57.52)22(54.3)(57.52)cosC

    STEP: Solve for the angle
    [−2 points ⇒ 0 / 5 points left]

    Now we can solve for the angle. We can expect to use the inverse cosine function since we will have to remove the cos from the angle.

    (61.76)2=(54.3)2+(57.52)22(54.3)(57.52)cosC3,814.2976=2,948.49+3,308.55046,246.672cosC3,814.2976=6,257.04046,246.672cosC2,442.7428=6,246.672cosC2,442.74286,246.672=cosC0.39104...=cosC

    As always, the inverse cosine operation comes with extra pieces: a second solution and a 360k° term. Altogether, that gives these solutions to the equation:

    0.39104...=cosCcos1(0.39104...)=cos1(cosC)

    This produces two solutions:

    C=66.98033...°+360k°,kZorC=66.98033...°+360k°,kZ

    These answers are all valid for the equation we solved. But the only valid answer for the question about the subtended angle is the positive one, because an angle in a triangle cannot be negative or greater than 180°. So angle C is 66.98033...°.

    Rounding to three decimal places, we get the final answer: 66.98°.


    Submit your answer as:
  2. If you are now told that the angle subtended at one of the flies by the other fly and the chameleon is 59°, determine the size of the angle subtended at the other fly by the other animals.

    INSTRUCTION: Round to three decimal places.
    Answer: The angle is °.
    one-of
    type(numeric.abserror(0.0004))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    This question tells you another angle. Label this angle on the diagram from Question 1. What can you do then?


    STEP: Use the sum of angles in a triangle
    [−1 point ⇒ 0 / 1 points left]

    This question sounds complicated, but it comes down to this: we now know another one of the angles in the triangle. We actually don't know which of the flies the angle is at - it could be either one. So let's shade one of the angles at the position of one of the flies.

    Now we know 2 of the angles in the triangle. The three angles add up to 180°, so:

    2nd flyangle at the=180°66.98033...°59°=54.01966...°54.02°
    NOTE: No flies were harmed in the making of this question.

    The angle is 54.02°.


    Submit your answer as:

ID is: 3568 Seed is: 9715

Predator and prey

Two hyenas are hunting a wildebeest in the tall grass. One of the hyenas is 47.29 m away from the wildebeest while the other hyena is 56.47 m away from the wildebeest. The hyenas are 64.25 m away from each other.

  1. What is the angle subtended by the two hyenas at the wildebeest's position?

    INSTRUCTION: Round to three decimal places.
    Answer: The angle is °.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by drawing a diagram. The shape you should use is a triangle. Then use the cosine rule to calculate the size of the angle you need.


    STEP: Draw a picture of the triangle
    [−1 point ⇒ 4 / 5 points left]

    We need to find an angle, but it is not obvious exactly which angle. We need to change the information in the question into a picture, so the first thing we need to do is draw a sketch. We should label it with everything we can.

    NOTE: The figure above is not drawn to scale. That is fine, but it means that we cannot draw any conclusions based on the appearance of the triangle. Only the labels are reliable.

    STEP: Identify the angle we want and write an equation using the cosine rule
    [−2 points ⇒ 2 / 5 points left]

    The question asks us to find the angle subtended by the two hyenas at the wildebeest's position. The word subtends means 'make an angle at a point from two other points.' The angle subtended at the wildebeest's position is the angle labelled wildebeest. This is because that angle is formed by the lines going to the two other animals.

    Let's use a letter W for wildebeest to represent the angle. Angle W is opposite from side 64.25 m, so using the cosine rule we get:

    (64.25)2=(47.29)2+(56.47)22(47.29)(56.47)cosW

    STEP: Solve for the angle
    [−2 points ⇒ 0 / 5 points left]

    Now we can solve for the angle. We can expect to use the inverse cosine function since we will have to remove the cos from the angle.

    (64.25)2=(47.29)2+(56.47)22(47.29)(56.47)cosW4,128.0625=2,236.3441+3,188.86095,340.9326cosW4,128.0625=5,425.2055,340.9326cosW1,297.1425=5,340.9326cosW1,297.14255,340.9326=cosW0.24286...=cosW

    As always, the inverse cosine operation comes with extra pieces: a second solution and a 360k° term. Altogether, that gives these solutions to the equation:

    0.24286...=cosWcos1(0.24286...)=cos1(cosW)

    This produces two solutions:

    W=75.94411...°+360k°,kZorW=75.94411...°+360k°,kZ

    These answers are all valid for the equation we solved. But the only valid answer for the question about the subtended angle is the positive one, because an angle in a triangle cannot be negative or greater than 180°. So angle W is 75.94411...°.

    Rounding to three decimal places, we get the final answer: 75.944°.


    Submit your answer as:
  2. If you are now told that the angle subtended at one of the hyenas by the other hyena and the wildebeest is 58.5°, determine the size of the angle subtended at the other hyena by the other animals.

    INSTRUCTION: Round to three decimal places.
    Answer: The angle is °.
    one-of
    type(numeric.abserror(0.0004))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    This question tells you another angle. Label this angle on the diagram from Question 1. What can you do then?


    STEP: Use the sum of angles in a triangle
    [−1 point ⇒ 0 / 1 points left]

    This question sounds complicated, but it comes down to this: we now know another one of the angles in the triangle. We actually don't know which of the hyenas the angle is at - it could be either one. So let's shade one of the angles at the position of one of the hyenas.

    Now we know 2 of the angles in the triangle. The three angles add up to 180°, so:

    2nd hyenaangle at the=180°75.94411...°58.5°=45.55588...°45.556°
    NOTE: The wildebeest was not harmed in the making of this question.

    The angle is 45.556°.


    Submit your answer as:

ID is: 3568 Seed is: 492

Predator and prey

A frog and two crickets are close to each other on the ground. The frog is watching the crickets, hoping for a tasty meal. One of the crickets is 55.37 cm away from the frog and the other cricket is 62.58 cm away from the frog. The crickets are 64.77 cm away from each other.

  1. What is the angle subtended by the two crickets at the frog's position?

    INSTRUCTION: Round to three decimal places.
    Answer: The angle is °.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by drawing a diagram. The shape you should use is a triangle. Then use the cosine rule to calculate the size of the angle you need.


    STEP: Draw a picture of the triangle
    [−1 point ⇒ 4 / 5 points left]

    We need to find an angle, but it is not obvious exactly which angle. We need to change the information in the question into a picture, so the first thing we need to do is draw a sketch. We should label it with everything we can.

    NOTE: The figure above is not drawn to scale. That is fine, but it means that we cannot draw any conclusions based on the appearance of the triangle. Only the labels are reliable.

    STEP: Identify the angle we want and write an equation using the cosine rule
    [−2 points ⇒ 2 / 5 points left]

    The question asks us to find the angle subtended by the two crickets at the frog's position. The word subtends means 'make an angle at a point from two other points.' The angle subtended at the frog's position is the angle labelled frog. This is because that angle is formed by the lines going to the two other animals.

    Let's use a letter F for frog to represent the angle. Angle F is opposite from side 64.77 cm, so using the cosine rule we get:

    (64.77)2=(55.37)2+(62.58)22(55.37)(62.58)cosF

    STEP: Solve for the angle
    [−2 points ⇒ 0 / 5 points left]

    Now we can solve for the angle. We can expect to use the inverse cosine function since we will have to remove the cos from the angle.

    (64.77)2=(55.37)2+(62.58)22(55.37)(62.58)cosF4,195.1529=3,065.8369+3,916.25646,930.1092cosF4,195.1529=6,982.09336,930.1092cosF2,786.9404=6,930.1092cosF2,786.94046,930.1092=cosF0.40214...=cosF

    As always, the inverse cosine operation comes with extra pieces: a second solution and a 360k° term. Altogether, that gives these solutions to the equation:

    0.40214...=cosFcos1(0.40214...)=cos1(cosF)

    This produces two solutions:

    F=66.28737...°+360k°,kZorF=66.28737...°+360k°,kZ

    These answers are all valid for the equation we solved. But the only valid answer for the question about the subtended angle is the positive one, because an angle in a triangle cannot be negative or greater than 180°. So angle F is 66.28737...°.

    Rounding to three decimal places, we get the final answer: 66.287°.


    Submit your answer as:
  2. If you are now told that the angle subtended at one of the crickets by the other cricket and the frog is 62.2°, determine the size of the angle subtended at the other cricket by the other animals.

    INSTRUCTION: Round to three decimal places.
    Answer: The angle is °.
    one-of
    type(numeric.abserror(0.0004))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    This question tells you another angle. Label this angle on the diagram from Question 1. What can you do then?


    STEP: Use the sum of angles in a triangle
    [−1 point ⇒ 0 / 1 points left]

    This question sounds complicated, but it comes down to this: we now know another one of the angles in the triangle. We actually don't know which of the crickets the angle is at - it could be either one. So let's shade one of the angles at the position of one of the crickets.

    Now we know 2 of the angles in the triangle. The three angles add up to 180°, so:

    2nd cricketangle at the=180°66.28737...°62.2°=51.51262...°51.513°
    NOTE: No crickets were harmed in the making of this question.

    The angle is 51.513°.


    Submit your answer as:

ID is: 4392 Seed is: 3216

Using the sin rule & cos rule to find unknown sides & angles

Adapted from DBE Nov 2016, Grade 11, P2, Q7
Maths formulas

Quadrilateral PQRS is drawn with QR=145 cm and RS=141 cm. It is also given that PS^Q=45.23°, SP^Q=105.48°, and QR^S=106.42°.

  1. Determine the length of QS.
  2. Determine the length of PQ.
INSTRUCTION: Round your answer to two decimal places, if necessary.
Answer:
  1. QS= cm
  2. PQ= cm
one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

In ΔSQR you want to determine a side and you know two sides and one angle. What rule can you use?


STEP: Use the cos rule in ΔSQR to find QS
[−3 points ⇒ 3 / 6 points left]
QS2=(RS)2+(QR)22(RS)(QR)(cosQR^S)=(141)2+(145)22(141)(145)(cos106.42°)=52,464.63...
QS=229.05159...229.05 cm

STEP: Use the sin rule in ΔPSQ to find PQ
[−3 points ⇒ 0 / 6 points left]
PQsinPS^Q=QSsinP^PQsin45.23°=229.05sin105.48°PQ=229.05×sin45.23°sin105.48°
PQ=168.73266...168.73 cm

Submit your answer as: and

ID is: 4392 Seed is: 5021

Using the sin rule & cos rule to find unknown sides & angles

Adapted from DBE Nov 2016, Grade 11, P2, Q7
Maths formulas

Quadrilateral JKLM is drawn with KL=175 mm and JK=119 mm. It is also given that JM^K=43.23°, MJ^K=106.35°, and LM=175 mm.

  1. Determine the length of KM.
  2. Determine the value of L^.
INSTRUCTION: Round your answer to two decimal places, if necessary.
Answer:
  1. KM= mm
  2. L^= °
one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

In ΔJMK you want to determine a side and you know two angles and one side. What rule can you use?


STEP: Use the sin rule in ΔJMK to find KM
[−3 points ⇒ 3 / 6 points left]
KMsinJ^=JKsinJM^KKMsin106.35°=119sin43.23°KM=119×sin106.35°sin43.23°
KM=166.71463...166.71 mm

STEP: Use the cos rule in ΔMKL to find angle L^
[−3 points ⇒ 0 / 6 points left]
KM2=(LM)2+(KL)22(LM)(KL)(cosL^)
2(175)(175)(cosL^)=(175)2+(175)2(166.71)2
cosL^=0.546249...
L^=56.88991...56.89°

Submit your answer as: and

ID is: 4392 Seed is: 5748

Using the sin rule & cos rule to find unknown sides & angles

Adapted from DBE Nov 2016, Grade 11, P2, Q7
Maths formulas

Quadrilateral PQRS is drawn with QR=129 mm and RS=132 mm. It is also given that PS^Q=50.30°, SP^Q=93.93°, and QR^S=53.35°.

  1. Determine the length of QS.
  2. Determine the length of PQ.
INSTRUCTION: Round your answer to two decimal places, if necessary.
Answer:
  1. QS= mm
  2. PQ= mm
one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

In ΔSQR you want to determine a side and you know two sides and one angle. What rule can you use?


STEP: Use the cos rule in ΔSQR to find QS
[−3 points ⇒ 3 / 6 points left]
QS2=(RS)2+(QR)22(RS)(QR)(cosQR^S)=(132)2+(129)22(132)(129)(cos53.35°)=13,736.11...
QS=117.20116...117.20 mm

STEP: Use the sin rule in ΔPSQ to find PQ
[−3 points ⇒ 0 / 6 points left]
PQsinPS^Q=QSsinP^PQsin50.30°=117.20sin93.93°PQ=117.20×sin50.30°sin93.93°
PQ=90.38616...90.39 mm

Submit your answer as: and

ID is: 3566 Seed is: 4715

Algebra and the cosine rule

Adapted from DBE Nov 2015, Grade 11, P2, Q7
Maths formulas

In the diagram, not drawn to scale, QR is the diameter of the circle. Triangle QPR is drawn with vertex P outside the circle. R^=x, QR=5d, PR=9d, and QP=8d.

  1. Determine the value of cosx.

    INSTRUCTION: Give your answer as a simplified fraction.
    Answer:

    cosx=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the cos rule in ΔQPR

    QP2=PR2+QR22(PR)(QR)cosx

    STEP: Use the cos rule
    [−1 point ⇒ 3 / 4 points left]

    We should focus on ΔQPR since x is an angle in this triangle, and we have expressions for all three sides. This means that we can use the cos rule.

    Write out the cos rule and substitute the given information:

    QP2=PR2+QR22(PR)(QR)cosx(8d)2=(9d)2+(5d)22(9d)(5d)cosx64d2=81d2+25d290d2cosx

    STEP: Make cosx the subject of your equation
    [−1 point ⇒ 2 / 4 points left]
    90d2cosx=81d2+25d264d2cosx=(81+2564)d290d2

    STEP: Simplify your expression
    [−2 points ⇒ 0 / 4 points left]
    cosx=42d290d2=715

    Submit your answer as:
  2. PR cuts the circumference of the circle at T.

    1. Determine the size of angle QT^R and select the correct reason.
    2. Determine QT in terms of x and d.
    Answer:
    1. QT^R= °
    2. QT=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Identify a right-angled triangle, and use a trigonometric ratio involving x and the side QT.


    STEP: Prove that ΔQTR is right-angled
    [−1 point ⇒ 2 / 3 points left]

    We are given that QR is the diameter of the circle.

    QT^R=90°( in semi circle)

    STEP: Write a trigonometric ratio involving x and the side QT
    [−2 points ⇒ 0 / 3 points left]

    We should focus on ΔQTR since we know that QR=5d and that the side QT and the angle x are in this triangle.

    In right-angled triangle ΔQTR, where QR^T=x:

    sinx=QTQRsinx=QT5dQT=5dsinx

    Submit your answer as: andand

ID is: 3566 Seed is: 3420

Algebra and the cosine rule

Adapted from DBE Nov 2015, Grade 11, P2, Q7
Maths formulas

In the diagram, not drawn to scale, AB is the diameter of the circle. Triangle ACB is drawn with vertex C outside the circle. B^=x, AB=5q, CB=8q, and AC=7q.

  1. Determine the value of cosx.

    INSTRUCTION: Give your answer as a simplified fraction.
    Answer:

    cosx=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the cos rule in ΔACB

    AC2=CB2+AB22(CB)(AB)cosx

    STEP: Use the cos rule
    [−1 point ⇒ 3 / 4 points left]

    We should focus on ΔACB since x is an angle in this triangle, and we have expressions for all three sides. This means that we can use the cos rule.

    Write out the cos rule and substitute the given information:

    AC2=CB2+AB22(CB)(AB)cosx(7q)2=(8q)2+(5q)22(8q)(5q)cosx49q2=64q2+25q280q2cosx

    STEP: Make cosx the subject of your equation
    [−1 point ⇒ 2 / 4 points left]
    80q2cosx=64q2+25q249q2cosx=(64+2549)q280q2

    STEP: Simplify your expression
    [−2 points ⇒ 0 / 4 points left]
    cosx=40q280q2=12

    Submit your answer as:
  2. CB cuts the circumference of the circle at T.

    1. Determine the size of angle AT^B and select the correct reason.
    2. Determine AT in terms of x and q.
    Answer:
    1. AT^B= °
    2. AT=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Identify a right-angled triangle, and use a trigonometric ratio involving x and the side AT.


    STEP: Prove that ΔATB is right-angled
    [−1 point ⇒ 2 / 3 points left]

    We are given that AB is the diameter of the circle.

    AT^B=90°( in semi circle)

    STEP: Write a trigonometric ratio involving x and the side AT
    [−2 points ⇒ 0 / 3 points left]

    We should focus on ΔATB since we know that AB=5q and that the side AT and the angle x are in this triangle.

    In right-angled triangle ΔATB, where AB^T=x:

    sinx=ATABsinx=AT5qAT=5qsinx

    Submit your answer as: andand

ID is: 3566 Seed is: 7106

Algebra and the cosine rule

Adapted from DBE Nov 2015, Grade 11, P2, Q7
Maths formulas

In the diagram, not drawn to scale, QR is the diameter of the circle. Triangle QPR is drawn with vertex P outside the circle. R^=x, QR=4a, PR=5a, and QP=4a.

  1. Determine the value of cosx.

    INSTRUCTION: Give your answer as a simplified fraction.
    Answer:

    cosx=

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the cos rule in ΔQPR

    QP2=PR2+QR22(PR)(QR)cosx

    STEP: Use the cos rule
    [−1 point ⇒ 3 / 4 points left]

    We should focus on ΔQPR since x is an angle in this triangle, and we have expressions for all three sides. This means that we can use the cos rule.

    Write out the cos rule and substitute the given information:

    QP2=PR2+QR22(PR)(QR)cosx(4a)2=(5a)2+(4a)22(5a)(4a)cosx16a2=25a2+16a240a2cosx

    STEP: Make cosx the subject of your equation
    [−1 point ⇒ 2 / 4 points left]
    40a2cosx=25a2+16a216a2cosx=(25+1616)a240a2

    STEP: Simplify your expression
    [−2 points ⇒ 0 / 4 points left]
    cosx=25a240a2=58

    Submit your answer as:
  2. PR cuts the circumference of the circle at T.

    1. Determine the size of angle QT^R and select the correct reason.
    2. Determine QT in terms of x and a.
    Answer:
    1. QT^R= °
    2. QT=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Identify a right-angled triangle, and use a trigonometric ratio involving x and the side QT.


    STEP: Prove that ΔQTR is right-angled
    [−1 point ⇒ 2 / 3 points left]

    We are given that QR is the diameter of the circle.

    QT^R=90°( in semi circle)

    STEP: Write a trigonometric ratio involving x and the side QT
    [−2 points ⇒ 0 / 3 points left]

    We should focus on ΔQTR since we know that QR=4a and that the side QT and the angle x are in this triangle.

    In right-angled triangle ΔQTR, where QR^T=x:

    sinx=QTQRsinx=QT4aQT=4asinx

    Submit your answer as: andand